{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{2}}+1=\\dfrac{1}{{{x}^{2}}}-4$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{ {\\pm \\dfrac{\\sqrt{5+\\sqrt{33}}}{2}} \\right\\}$","B. $S=\\left\\{ {\\pm \\dfrac{\\sqrt{-5+\\sqrt{33}}}{2}} \\right\\}$","C. $S=\\left\\{ {\\dfrac{5 \\pm\\sqrt{33}}{4}} \\right\\}$","D. $S=\\left\\{ { \\dfrac{-5\\pm\\sqrt{33}}{4}} \\right\\}$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 2: Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng<br\/>B\u01b0\u1edbc 3: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh thu \u0111\u01b0\u1ee3c. So s\u00e1nh nghi\u1ec7m v\u1edbi \u0111i\u1ec1u ki\u1ec7n v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $x\\ne 0$ . Khi \u0111\u00f3<br\/>$\\begin{align} & 2{{x}^{2}}+1=\\dfrac{1}{{{x}^{2}}}-4 \\\\ & \\Leftrightarrow 2{{x}^{4}}+5{{x}^{2}}-1=0 \\\\ \\end{align}$<br\/>\u0110\u1eb7t $t={{x}^{2}},t\\ge 0.$ Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{align} & 2{{t}^{2}}+5t-1=0 \\\\ & \\Delta =25+4.2=33 \\\\ \\end{align}$ <br\/>Suy ra ${{t}_{1}}=\\dfrac{-5+\\sqrt{33}}{4}$ (th\u1ecfa m\u00e3n) v\u00e0 ${{t}_{2}}=\\dfrac{-5-\\sqrt{33}}{4}$ (lo\u1ea1i)<br\/>V\u1edbi ${{t}_{1}}=\\dfrac{-5+\\sqrt{33}}{4}$ suy ra ${{x}^{2}}=\\dfrac{-5+\\sqrt{33}}{4}\\Leftrightarrow x=\\pm \\dfrac{\\sqrt{-5+\\sqrt{33}}}{2}$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ {\\pm \\dfrac{\\sqrt{-5+\\sqrt{33}}}{2}} \\right\\}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":941},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $4\\sqrt{3-x}+6=5x$ (1)<br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x =$ _input_ ","hint":"\u00c1p d\u1ee5ng m\u1ed9t trong hai ph\u01b0\u01a1ng ph\u00e1p:<br\/>+ \u0110\u1eb7t \u1ea9n ph\u1ee5 (t\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n ph\u1ee5)<br\/>+ \u0110\u1eb7t \u0111i\u1ec1u ki\u1ec7n r\u1ed3i b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf khi hai v\u1ebf \u0111\u1ec1u kh\u00f4ng \u00e2m.<br\/><b>Ch\u00fa \u00fd:<\/b> Sau khi \u0111\u00e3 t\u00ecm \u0111\u01b0\u1ee3c nghi\u1ec7m c\u1ea7n ph\u1ea3i ki\u1ec3m tra l\u1ea1i \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ch\u1ecdn nghi\u1ec7m th\u00edch h\u1ee3p.","explain":"<span class='basic_left'><b> C\u00e1ch 1: <\/b><br\/> \u0110i\u1ec1u ki\u1ec7n: $x\\le 3$ . Khi \u0111\u00f3:<br\/>$\\left( 1 \\right)\\Leftrightarrow 4\\sqrt{3-x}+5\\left( 3-x \\right)-9=0$ <br\/>\u0110\u1eb7t $y=\\sqrt{3-x},y\\ge 0$, ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$4y+5{{y}^{2}}-9=0$<br\/>$\\Leftrightarrow 5y^2+4y-9=0$<br\/>Ta c\u00f3: $a+b+c=5+4-9=0$<br\/>$\\Rightarrow \\left[ \\begin{aligned} & y=1\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & y=-\\dfrac{9}{5} \\,(\\text{lo\u1ea1i})\\\\ \\end{aligned} \\right.$<br\/>+ V\u1edbi $y=1$$\\Leftrightarrow\\sqrt{3-x}=1\\Leftrightarrow 3-x=1\\Leftrightarrow x=2$ (th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n $x\\le 3$)<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=2$ <br\/><b>C\u00e1ch 2:<\/b><br\/> \u0110i\u1ec1u ki\u1ec7n: $x\\le 3$<br\/>T\u00e1ch ri\u00eang c\u0103n th\u1ee9c \u1edf m\u1ed9t v\u1ebf, ta \u0111\u01b0\u1ee3c:<br\/>$4\\sqrt{3-x}=5x-6$ <br\/>V\u1edbi \u0111i\u1ec1u ki\u1ec7n $ 5x-6 \\ge 0 $ t\u1ee9c $x \\ge \\dfrac{6}{5},$ b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & 16\\left( 3-x \\right)={{\\left( 5x-6 \\right)}^{2}} \\\\ & \\Leftrightarrow 48-16x=25{{x}^{2}}-60x+36 \\\\ & \\Leftrightarrow 25{{x}^{2}}-44x-12=0 \\\\ \\end{aligned}$ <br\/>$\\Delta '={{22}^{2}}+25.12=784\\Rightarrow \\sqrt{\\Delta }=28$ <br\/>Suy ra ${{x}_{1}}=\\dfrac{22+28}{25}=2$ (th\u1ecfa m\u00e3n $x\\le 3$ v\u00e0 $x\\ge \\dfrac{6}{5}$)<br\/>${{x}_{2}}=\\dfrac{22-28}{25}=-\\dfrac{6}{25}$ (lo\u1ea1i v\u00ec kh\u00f4ng th\u1ecfa m\u00e3n $x\\ge \\dfrac{6}{5})$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=2$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $2.$ <br\/><\/span><span class='basic_green'>L\u01b0u \u00fd:<\/span> V\u1edbi c\u00e1ch gi\u1ea3i 2:<br\/>N\u1ebfu kh\u00f4ng \u0111\u1eb7t \u0111i\u1ec1u ki\u1ec7n $5x-6\\ge 0$ th\u00ec sau khi t\u00ecm \u0111\u01b0\u1ee3c nghi\u1ec7m, ta ph\u1ea3i th\u1eed l\u1ea1i \u0111\u1ec3 ch\u1ecdn ra nghi\u1ec7m. C\u00e1ch l\u00e0m n\u00e0y ph\u1ee9c t\u1ea1p h\u01a1n so v\u1edbi vi\u1ec7c \u0111\u1eb7t \u0111i\u1ec1u ki\u1ec7n $x\\ge \\dfrac{6}{5}$ r\u1ed3i \u0111\u1ed1i chi\u1ebfu c\u00e1c nghi\u1ec7m v\u1edbi \u0111i\u1ec1u ki\u1ec7n.<br\/>C\u00e1ch gi\u1ea3i t\u1ed5ng qu\u00e1t c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $\\sqrt{f\\left( x \\right)}=g\\left( x \\right)$ <br\/>\u0110i\u1ec1u ki\u1ec7n: $f\\left( x \\right)\\ge 0$<br\/> $\\sqrt{f\\left( x \\right)}=g\\left( x \\right)\\Leftrightarrow \\left\\{ \\begin{aligned} & g\\left( x \\right)\\ge 0 \\\\ & f\\left( x \\right)={{\\left[ g\\left( x \\right) \\right]}^{2}} \\\\ \\end{aligned} \\right.$<\/span>"}]}],"id_ques":942},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $6+\\sqrt{{{x}^{2}}-7x+12}=2x$ (1) <br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x=$_input_","hint":"C\u00f4 l\u1eadp c\u0103n th\u1ee9c \u1edf m\u1ed9t v\u1ebf","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: ${{x}^{2}}-7x+12\\ge 0$ (2)<br\/>$\\left( 1 \\right)\\Leftrightarrow \\sqrt{{{x}^{2}}-7x+12}=2x-6$ <br\/>V\u1edbi \u0111i\u1ec1u ki\u1ec7n $x\\ge 3\\,\\,\\,(3)$, b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf, ta c\u00f3: <br\/>$\\begin{align} & {{x}^{2}}-7x+12=4{{x}^{2}}-24x+36 \\\\ & \\Leftrightarrow 3{{x}^{2}}-17x+24=0 \\\\ \\end{align}$ <br\/>$\\Delta ={{17}^{2}}-4.3.24=1$ <br\/>Suy ra:<br\/> ${{x}_{1}}=\\dfrac{17+1}{6}=3$ (th\u1ecfa m\u00e3n (2) v\u00e0 (3)); ${{x}_{2}}=\\dfrac{17-1}{6}=\\dfrac{8}{3}$ (kh\u00f4ng th\u1ecfa m\u00e3n (3))<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=3$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3.$<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>Khi \u0111\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ngh\u0129a, n\u1ebfu \u0111i\u1ec1u ki\u1ec7n ph\u1ee9c t\u1ea1p th\u00ec ta ch\u1ec9 c\u1ea7n \u0111\u1eb7t \u0111i\u1ec1u ki\u1ec7n m\u00e0 kh\u00f4ng nh\u1ea5t thi\u1ebft ph\u1ea3i gi\u1ea3i \u0111i\u1ec1u ki\u1ec7n \u0111\u00f3. Sau khi t\u00ecm \u0111\u01b0\u1ee3c nghi\u1ec7m, ta th\u1eed l\u1ea1i xem nghi\u1ec7m \u0111\u00f3 c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n hay kh\u00f4ng.<\/span>"}]}],"id_ques":943},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-5x+13=4\\sqrt{{{x}^{2}}-5x+9}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 ","select":["A. $S=\\left\\{ 5 \\pm \\sqrt{5} \\right\\}$","B. $S=\\left\\{ \\dfrac{-5\\pm \\sqrt{5}}{2} \\right\\}$","C. $S=\\left\\{ \\dfrac{5\\pm \\sqrt{5}}{2} \\right\\}$"],"hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 $t=\\sqrt{{{x}^{2}}-5x+9}$","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: ${{x}^{2}}-5x+9\\ge 0\\Leftrightarrow {{\\left( x-\\dfrac{5}{2} \\right)}^{2}}+\\dfrac{11}{4}\\ge 0$ (lu\u00f4n \u0111\u00fang v\u1edbi m\u1ecdi $x$)<br\/>\u0110\u1eb7t $t=\\sqrt{{{x}^{2}}-5x+9},$ $t\\ge 0$. Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned} & {{t}^{2}}+4=4t \\\\ & \\Leftrightarrow {{\\left( t-2 \\right)}^{2}}=0 \\\\ & \\Leftrightarrow t=2 (\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned}$<br\/> Suy ra $\\sqrt{{{x}^{2}}-5x+9}=2$ <br\/>$\\begin{aligned} & \\Leftrightarrow {{x}^{2}}-5x+9=4 \\\\ & \\Leftrightarrow {{x}^{2}}-5x+5=0 \\\\ \\end{aligned}$<br\/>$\\Delta =25-20=5$<br\/> Suy ra $x=\\dfrac{5\\pm \\sqrt{5}}{2}$<br\/> V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ \\dfrac{5\\pm \\sqrt{5}}{2} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<br\/> <\/span><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span> <br\/>\u1ede b\u00e0i n\u00e0y, ta kh\u00f4ng n\u00ean b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf v\u00ec khi \u0111\u00f3, qu\u00e1 tr\u00ecnh bi\u1ebfn \u0111\u1ed5i ph\u1ee9c t\u1ea1p h\u01a1n r\u1ea5t nhi\u1ec1u v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh thu \u0111\u01b0\u1ee3c s\u1ebd c\u00f3 b\u1eadc 4.<\/span>","column":3}]}],"id_ques":944},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-1"],["3"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\sqrt{3x+7}-\\sqrt{x+1}=2$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{$_input_;_input_$\\}$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 2: \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $\\sqrt{f(x)}=\\sqrt{g(x)}+b, b>0$ sau \u0111\u00f3 b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf.<br\/>B\u01b0\u1edbc 3: Bi\u1ebfn \u0111\u1ed5i v\u00e0 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng: $\\sqrt{f(x)}=m$. <br\/>B\u01b0\u1edbc 4: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh thu \u0111\u01b0\u1ee3c, so s\u00e1nh nghi\u1ec7m v\u1edbi \u0111i\u1ec1u ki\u1ec7n v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $\\left\\{ \\begin{aligned} & 3x+7\\ge 0 \\\\ & x+1\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge -\\dfrac{7}{3} \\\\ & x\\ge -1 \\\\ \\end{aligned} \\right.\\Leftrightarrow x\\ge -1$<br\/> $\\left( 1 \\right)\\Leftrightarrow \\sqrt{3x+7}=2+\\sqrt{x+1}$ <br\/>B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & 3x+7=4+4\\sqrt{x+1}+x+1 \\\\ & \\Leftrightarrow 2\\sqrt{x+1}=x+1 \\\\ & \\Leftrightarrow 4\\left( x+1 \\right)={{\\left( x+1 \\right)}^{2}} \\\\ & \\Leftrightarrow {{x}^{2}}-2x-3=0 \\\\ \\end{aligned}$ <br\/>$\\Leftrightarrow \\left[ \\begin{aligned} & x=-1 \\\\ & x=3 \\\\ \\end{aligned} \\right.$(th\u1ecfa m\u00e3n)<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ -1;3 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1;$ $3.$<br\/><\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a c\u0103n th\u1ee9c d\u1ea1ng $\\sqrt{f(x)}+\\sqrt{g(x)}=\\sqrt{h(x)}+\\sqrt{k(x)}$<br\/>B\u01b0\u1edbc 1: \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $f(x) \\ge 0;$ $g(x) \\ge 0;$ $h(x)\\ge 0;$ $k(x)\\ge 0$<br\/>B\u01b0\u1edbc 2: B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh<br\/><b>Ch\u00fa \u00fd:<\/b> Ch\u1ec9 b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf khi c\u1ea3 hai v\u1ebf \u0111\u1ec1u kh\u00f4ng \u00e2m. N\u1ebfu ta kh\u00f4ng \u0111\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 hai v\u1ebf kh\u00f4ng \u00e2m th\u00ec sau khi t\u00ecm \u0111\u01b0\u1ee3c nghi\u1ec7m, ta ph\u1ea3i thay nghi\u1ec7m v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u \u0111\u1ec3 th\u1eed l\u1ea1i.<\/span>"}]}],"id_ques":945},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $2\\left| x-3 \\right|={{x}^{2}}-5$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 ","select":["A. $S=\\left\\{ 1\\pm \\sqrt{3} \\right\\}$","B. $S=\\left\\{ -1\\pm \\sqrt{3} \\right\\}$","C. $S=\\left\\{ 1\\pm 2\\sqrt{3} \\right\\}$","D. $S=\\left\\{ -1\\pm 2\\sqrt{3} \\right\\}$"],"hint":"B\u1ecf gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i trong ph\u01b0\u01a1ng tr\u00ecnh b\u1eb1ng \u0111\u1ecbnh ngh\u0129a:<br\/> $\\left| A \\right|=\\left\\{ \\begin{aligned} & A\\,\\,\\,\\,\\,\\,\\,\\left(\\text{n\u1ebfu}\\, A\\ge 0 \\right) \\\\ & -A\\,\\,\\,\\,\\,\\left(\\text{n\u1ebfu}\\, A<0 \\right) \\\\ \\end{aligned} \\right.$","explain":"<span class='basic_left'>X\u00e9t hai tr\u01b0\u1eddng h\u1ee3p:<br\/>+ V\u1edbi $x\\ge 3$ th\u00ec $\\left| x-3 \\right|=x-3,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\begin{aligned} & 2\\left( x-3 \\right)={{x}^{2}}-5 \\\\ & \\Leftrightarrow {{x}^{2}}-2x+1=0 \\\\ & \\Leftrightarrow {{\\left( x-1 \\right)}^{2}}=0 \\\\ & \\Leftrightarrow x=1 \\,(\\text{lo\u1ea1i}) \\\\ \\end{aligned}$ <br\/>+ V\u1edbi $x<3$ th\u00ec $\\left| x-3 \\right|=3-x,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/> $\\begin{aligned} & 2\\left( 3-x \\right)={{x}^{2}}-5 \\\\ & \\Leftrightarrow {{x}^{2}}+2x-11=0 \\\\ \\end{aligned}$ <br\/>$\\Delta '=1+11=12\\Rightarrow \\sqrt{\\Delta '}=2\\sqrt{3}$ <br\/>Suy ra ${{x}_{1}}=-1+2\\sqrt{3};$ ${{x}_{2}}=-1-2\\sqrt{3}$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ -1\\pm 2\\sqrt{3} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<br\/> <\/span><\/span>","column":2}]}],"id_ques":946},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["0"],["2"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+\\left| x-1 \\right|=2x+1$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{$_input_;_input_$\\}$","hint":"","explain":"<span class='basic_left'>X\u00e9t hai tr\u01b0\u1eddng h\u1ee3p:<br\/>+ V\u1edbi $x\\ge 1$ th\u00ec $\\left| x-1 \\right|=x-1,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>${{x}^{2}}+x-1=2x+1$<br\/>$\\Leftrightarrow {{x}^{2}}-x-2=0$<br\/>$\\Leftrightarrow \\left[ \\begin{align} & x=-1 \\\\ & x=2 \\\\ \\end{align} \\right.$ (do $a-b+c=0$)<br\/>$\\Leftrightarrow x=2$ (do $x\\ge 1$)<br\/>+ V\u1edbi $x<1$ th\u00ec $\\left| x-1 \\right|=-x+1,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\begin{aligned} & {{x}^{2}}-x+1=2x+1 \\\\ & \\Leftrightarrow {{x}^{2}}-3x=0 \\\\ & \\Leftrightarrow x\\left( x-3 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0\\,\\,\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ & x=3\\,\\,\\,\\,\\,(\\text {lo\u1ea1i}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 0;2 \\right\\}.$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0;$ $2.$<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span> C\u00e1ch 2:<br\/>\u0110\u1eb7t $t=\\left| x-1 \\right|.$ \u0110i\u1ec1u ki\u1ec7n $t\\ge 0.$ <br\/>Ta c\u00f3 ${{t}^{2}}={{\\left| x-1 \\right|}^{2}}={{x}^{2}}-2x+1\\Rightarrow {{x}^{2}}-2x={{t}^{2}}-1$ .<br\/>Khi \u0111\u00f3 $\\left( 1 \\right)\\Leftrightarrow {{t}^{2}}-1+t-1=0\\Leftrightarrow {{t}^{2}}+t-2=0\\Leftrightarrow \\left[ \\begin{aligned} & t=1 \\\\ & t=-2\\,(\\text{lo\u1ea1i}) \\\\ \\end{aligned} \\right.$ <br\/>V\u1edbi $t=1,$ ta c\u00f3 $\\left| x-1 \\right|=1\\Leftrightarrow \\left[ \\begin{aligned} & x-1=1 \\\\ & x-1=-1 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=2 \\\\ & x=0 \\\\ \\end{aligned} \\right.$ <\/span>"}]}],"id_ques":947},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"C\u00e1c ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 \u0111\u01b0\u1ee3c \u0111\u00e1nh s\u1ed1 th\u1ee9 t\u1ef1, ta ch\u1ec9 vi\u1ec7c \u0111i\u1ec1n theo s\u1ed1 th\u1ee9 t\u1ef1 \u1edf \u0111\u1ec1 b\u00e0i<br\/>V\u00ed d\u1ee5: $1$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1), (2), (3) th\u00ec ta \u0111i\u1ec1n 1;2;3","temp":"fill_the_blank","correct":[[["1;5","5;1"],["3;4","4;3"]]],"list":[{"point":5,"width":70,"content":"","type_input":"","ques":"<span class='basic_left'>Cho c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh:<br\/> $\\begin{aligned} & {{x}^{3}}+3{{x}^{2}}+x-5=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{x}^{3}}-4{{x}^{2}}+5x-6=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ & {{x}^{3}}-4{{x}^{2}}+x+6=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 3 \\right) \\\\ & {{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}-5x-2=0\\,\\,\\,\\,\\,\\left( 4 \\right) \\\\ & {{x}^{4}}-{{x}^{3}}+6{{x}^{2}}+x-7=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 5 \\right) \\\\ \\end{aligned}$<br\/>$1$ l\u00e0 nghi\u1ec7m c\u1ee7a c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh: _input_<br\/>$-1$ l\u00e0 nghi\u1ec7m c\u1ee7a c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh:_input_ <\/span>","hint":"Cho ph\u01b0\u01a1ng tr\u00ecnh $f \\left( x \\right)=0$ trong \u0111\u00f3 $f(x)$ l\u00e0 \u0111a th\u1ee9c v\u1edbi bi\u1ebfn $x.$<br\/>+ N\u1ebfu t\u1ed5ng c\u00e1c h\u1ec7 s\u1ed1 c\u1ee7a $f(x)$ b\u1eb1ng $0$ th\u00ec $1$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>+ N\u1ebfu t\u1ed5ng c\u00e1c h\u1ec7 s\u1ed1 c\u1ee7a c\u00e1c h\u1ea1ng t\u1eed b\u1eadc ch\u1eb5n b\u1eb1ng t\u1ed5ng c\u00e1c h\u1ec7 s\u1ed1 c\u1ee7a c\u00e1c h\u1ea1ng t\u1eed b\u1eadc l\u1ebb c\u1ee7a $f(x)$ th\u00ec $-1$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. ","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3: $a+b+c+d=1+3+1-5=0$ n\u00ean $1$ l\u00e0 nghi\u1ec7m c\u1ee7a (1)<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (2) kh\u00f4ng c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n n\u00e0o trong hai \u0111i\u1ec1u ki\u1ec7n tr\u00ean n\u00ean kh\u00f4ng c\u00f3 nghi\u1ec7m l\u00e0 $1$ v\u00e0 $-1$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (3) c\u00f3 $a+c=b+d=2$ n\u00ean $-1$ l\u00e0 nghi\u1ec7m c\u1ee7a (3)<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (4) c\u00f3 $a+c+e=b+d=-3$ n\u00ean $-1$ l\u00e0 nghi\u1ec7m c\u1ee7a (4)<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (5) c\u00f3: $a+b+c+d=1-1+6+1-7=0$ n\u00ean $1$ l\u00e0 nghi\u1ec7m c\u1ee7a (5)<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng d\u00f2ng 1 l\u00e0 1;5 ho\u1eb7c 5;1 <br\/>\u0110\u00e1p \u00e1n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng d\u00f2ng 2 l\u00e0 3;4 ho\u1eb7c 4;3 <\/span><\/span>"}]}],"id_ques":948},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1"],["2"],["3"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{3}}-6{{x}^{2}}+11x-6=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{$_input_;_input_;_input_$\\}$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc ba $a{{x}^{3}}+b{{x}^{2}}+cx+d=0\\,\\,\\,\\left( a\\ne 0 \\right)$ bi\u1ebft m\u1ed9t nghi\u1ec7m $x=\\alpha $ <br\/>B\u01b0\u1edbc 1: Nh\u1ea9m nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. B\u1eb1ng ph\u00e9p chia \u0111a th\u1ee9c (ho\u1eb7c d\u00f9ng l\u01b0\u1ee3c \u0111\u1ed3 Horner), ta ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed: $\\left( x-\\alpha \\right)\\left( a{{x}^{2}}+{{b}_{1}}x+{{c}_{1}} \\right)$ <br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00edch \u1edf b\u01b0\u1edbc 1 v\u00e0 k\u1ebft lu\u1eadn t\u1eadp nghi\u1ec7m.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: $a+b+c+d=1-6+11-6=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=1$<br\/>Th\u1ef1c hi\u1ec7n ph\u00e9p chia \u0111a th\u1ee9c \u1edf v\u1ebf tr\u00e1i c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh cho $x-1,$ ta \u0111\u01b0\u1ee3c th\u01b0\u01a1ng $x^2-5x+6.$<br\/>Khi \u0111\u00f3 ${{x}^{3}}-6{{x}^{2}}+11x-6=0$<br\/>$\\begin{aligned} & \\Leftrightarrow \\left( x-1 \\right)\\left( {{x}^{2}}-5x+6 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-1=0 \\\\ & {{x}^{2}}-5x+6=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=1 \\\\ & {{x}^{2}}-5x+6=0 \\\\ \\end{aligned} \\right.\\\\ \\end{aligned}$<br\/> Ph\u01b0\u01a1ng tr\u00ecnh $x^2-5x+6=0$ c\u00f3 $\\Delta =25-24=1>0.$<br\/> Suy ra ${{x}_{1}}=\\dfrac{5+1}{2}=3;{{x}_{2}}=\\dfrac{5-1}{2}=2$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 1;2;3 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1;$ $2;$ $3.$<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/><b>Ph\u00e2n t\u00edch \u0111a th\u1ee9c b\u1eb1ng l\u01b0\u1ee3c \u0111\u1ed3 Horner: <\/b><br\/>Gi\u1ea3 s\u1eed chia \u0111a th\u1ee9c $P\\left( x \\right)={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+...+{{a}_{n-1}}x+{{a}_{n}}$ cho $x-\\alpha $ , ta \u0111\u01b0\u1ee3c: <br\/>$P\\left( x \\right)=\\left( x-\\alpha \\right)\\left( {{b}_{0}}{{x}^{n-1}}+{{b}_{1}}{{x}^{n-2}}+...+{{b}_{n-1}} \\right)+{{b}_{n}}$ <br\/>Khi \u0111\u00f3, ta c\u00f3 s\u01a1 \u0111\u1ed3 x\u00e1c \u0111\u1ecbnh $b_i: $<br\/><table><tr><td>H\u1ec7 s\u1ed1 c\u1ee7a $P\\left( x \\right)$ <\/td><td>${{a}_{0}}$ <\/td><td>${{a}_{1}}$ <\/td><td>${{a}_{2}}$ <\/td><td>...<\/td><td>${{a}_{n}}$ <\/td><\/tr><tr><td>$\\alpha $ <\/td><td>${{b}_{0}}$ <\/td><td>${{b}_{1}}$ <\/td><td>${{b}_{2}}$ <\/td><td>... <\/td><td>${{b}_{n}}$ <\/td><\/tr><\/table><br\/>D\u00f2ng 2 l\u00e0 h\u1ec7 s\u1ed1 c\u1ee7a th\u01b0\u01a1ng, trong \u0111\u00f3 ${{b}_{0}}={{a}_{0}};{{b}_{1}}=\\alpha {{b}_{0}}+{{a}_{1}};{{b}_{2}}=\\alpha {{b}_{1}}+{{a}_{2}};...;{{a}_{n}}=\\alpha {{b}_{n-1}}+{{a}_{n}}$<br\/> Ta c\u00f3 l\u01b0\u1ee3c \u0111\u1ed3 Horner cho \u0111a th\u1ee9c $P\\left( x \\right)={{x}^{3}}-6{{x}^{2}}+11x-6$<br\/><table><tr><td>H\u1ec7 s\u1ed1 c\u1ee7a $P\\left( x \\right)$ <\/td><td>$1$ <\/td><td>$-6$ <\/td><td>$11$ <\/td><td>$-6$<\/td><\/tr><tr><td>$\\alpha =1$ <\/td><td>$1$ <\/td><td>$-5$ <\/td><td>$6$ <\/td><td>$0$ <\/td><\/tr><\/table><br\/>Khi \u0111\u00f3 $P\\left( x \\right)=\\left( x-1 \\right)\\left( {{x}^{2}}-5x+6 \\right)$ <br\/><b> L\u01b0u \u00fd:<\/b> Trong qu\u00e1 tr\u00ecnh gi\u1ea3i b\u00e0i to\u00e1n, vi\u1ec7c ph\u00e2n t\u00edch \u0111a th\u1ee9c b\u1eb1ng c\u00e1ch chia \u0111a th\u1ee9c hay d\u00f9ng l\u01b0\u1ee3c \u0111\u1ed3 Horner, ta c\u00f3 th\u1ec3 th\u1ef1c hi\u1ec7n \u1edf nh\u00e1p m\u00e0 kh\u00f4ng nh\u1ea5t thi\u1ebft ph\u1ea3i tr\u00ecnh b\u00e0y v\u00e0o l\u1eddi gi\u1ea3i.<\/span>"}]}],"id_ques":949},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Bi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh $6{{x}^{3}}-7{{x}^{2}}-16x+m=0$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $2.$<br\/>C\u00e1c nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: ","select":["A. $x=\\dfrac{2}{3}$ v\u00e0 $x=-\\dfrac{3}{2}$ ","B. $x=-\\dfrac{2}{3}$ v\u00e0 $x=-\\dfrac{3}{2}$ ","C. $x=-\\dfrac{2}{3}$ v\u00e0 $x=\\dfrac{3}{2}$","D. $x=\\dfrac{4}{3}$ v\u00e0 $x=-3$"],"hint":"T\u00ecm $m$ tr\u01b0\u1edbc sau \u0111\u00f3 d\u00f9ng l\u01b0\u1ee3c \u0111\u1ed3 Horner \u0111\u1ec3 ph\u00e2n t\u00edch v\u1ebf tr\u00e1i ph\u01b0\u01a1ng tr\u00ecnh th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh $6{{x}^{3}}-7{{x}^{2}}-16x+m=0$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $2$ n\u00ean<br\/> $6.2^3-7.2^2-16.2 +m=0$<br\/>$\\begin{aligned} & \\Leftrightarrow -12+m=0 \\\\ & \\Leftrightarrow m=12 \\\\ \\end{aligned}$ <br\/>V\u1edbi $m=12,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $6{{x}^{3}}-7{{x}^{2}}-16x+12=0$ (1)<br\/>L\u01b0\u1ee3c \u0111\u1ed3 Horner:<br\/><table><tr><td>H\u1ec7 s\u1ed1<\/td><td>$6$ <\/td><td>$-7$ <\/td><td>$-16$ <\/td><td>$12$<\/td><\/tr><tr><td>$x=2 $ <\/td><td>$6$ <\/td><td>$5$ <\/td><td>$-6$ <\/td><td>$0$ <\/td><\/tr><\/table><br\/> Khi \u0111\u00f3<br\/>$\\begin{aligned} & \\left( 1 \\right)\\Leftrightarrow \\left( x-2 \\right)\\left( 6{{x}^{2}}+5x-6 \\right)=0 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow \\left[ \\begin{aligned} & x-2=0 \\\\ & 6{{x}^{2}}+5x-6=0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh $6{{x}^{2}}+5x-6=0$ c\u00f3 $\\Delta =169$ <br\/>Suy ra ${{x}_{1}}=\\dfrac{-5+13}{12}=\\dfrac{2}{3}$ ; ${{x}_{2}}=\\dfrac{-5-13}{12}=-\\dfrac{3}{2}$ <br\/>V\u1eady v\u1edbi $m=12$ th\u00ec nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x=\\dfrac{2}{3}$ v\u00e0 $x=-\\dfrac{3}{2}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A <\/span><\/span>","column":2}]}],"id_ques":950},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"X\u00e1c \u0111\u1ecbnh $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh sau c\u00f3 ba nghi\u1ec7m ph\u00e2n bi\u1ec7t:<br\/> ${{x}^{3}}+\\left( 2m-3 \\right){{x}^{2}}+\\left( {{m}^{2}}-2m+2 \\right)x-{{m}^{2}}=0$ (1)","select":["A. $m>\\dfrac{1}{2}$ ","B. $m<\\dfrac{1}{2}$ ","C. $m<\\dfrac{1}{2}$v\u00e0 $m\\ne 1\\pm \\sqrt{2}$","D. $m<\\dfrac{1}{2}$v\u00e0 $m\\ne -1\\pm \\sqrt{2}$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Nh\u1ea9m nghi\u1ec7m sau \u0111\u00f3 d\u00f9ng l\u01b0\u1ee3c \u0111\u1ed3 Horner ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed: $(x-\\alpha)f(x)$<br\/>B\u01b0\u1edbc 2:\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 ba nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh $f(x)=0$ c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t kh\u00e1c $\\alpha$. Gi\u1ea3i \u0111i\u1ec1u ki\u1ec7n \u0111\u00f3 \u0111\u1ec3 t\u00ecm $m$<br\/><b>Ch\u00fa \u00fd:<\/b> Ph\u01b0\u01a1ng tr\u00ecnh $f(x)=0$ c\u00f3 nghi\u1ec7m kh\u00e1c $\\alpha \\Leftrightarrow f(\\alpha) \\ne 0 $<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $a+b+c+d=1+(2m-3)+(m^2-2m+2)-m^2=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=1.$<br\/>Ta c\u00f3 l\u01b0\u1ee3c \u0111\u1ed3 Horner:<br\/><table><tr><td>H\u1ec7 s\u1ed1<\/td><td>$1$ <\/td><td>$2m-3$ <\/td><td>$m^2-2m+2$ <\/td><td>$-m^2$<\/td><\/tr><tr><td>$x=1 $ <\/td><td>$1$ <\/td><td>$2m-2=2(m-1)$ <\/td><td>$m^2$ <\/td><td>$0$ <\/td><\/tr><\/table><br\/> Khi \u0111\u00f3<br\/> $\\begin{aligned} & \\left( 1 \\right)\\Leftrightarrow \\left( x-1 \\right)\\left[ {{x}^{2}}+2\\left( m-1 \\right)x+{{m}^{2}} \\right]=0 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow \\left[ \\begin{aligned} & x=1 \\\\ & {{x}^{2}}+2\\left( m-1 \\right)x+{{m}^{2}}=0\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 ba nghi\u1ec7m ph\u00e2n bi\u1ec7t th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t kh\u00e1c $1$<br\/>$\\begin{aligned} & \\Leftrightarrow \\left\\{ \\begin{aligned} & \\Delta '={{\\left( m-1 \\right)}^{2}}-{{m}^{2}}>0 \\\\ & {{1}^{2}}+2\\left( m-1 \\right).1+{{m}^{2}}\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & -2m+1>0 \\\\ & {{m}^{2}}+2m-1\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & m<\\dfrac{1}{2} \\\\ & m\\ne -1\\pm \\sqrt{2} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span><\/span>","column":2}]}],"id_ques":951},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-4"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{7}{{{x}^{2}}-1}+\\dfrac{8x}{{{x}^{2}}-2x+1}=\\dfrac{25-9x}{{{x}^{3}}-{{x}^{2}}-x+1}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $x=$_input_","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 2: Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c hai v\u1ebf r\u1ed3i kh\u1eed m\u1eabu th\u1ee9c<br\/>B\u01b0\u1edbc 3: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1eeba nh\u1eadn \u0111\u01b0\u1ee3c<br\/>B\u01b0\u1edbc 4: Trong c\u00e1c gi\u00e1 tr\u1ecb t\u00ecm \u0111\u01b0\u1ee3c c\u1ee7a \u1ea9n, lo\u1ea1i c\u00e1c gi\u00e1 tr\u1ecb kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh. C\u00e1c gi\u00e1 tr\u1ecb th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: ${{x}^{3}}-{{x}^{2}}-x+1={{x}^{2}}\\left( x-1 \\right)-\\left( x-1 \\right)=\\left( {{x}^{2}}-1 \\right)\\left( x-1 \\right)={{\\left( x-1 \\right)}^{2}}\\left( x+1 \\right)$<br\/>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\pm 1$<br\/>Khi \u0111\u00f3<br\/>$\\begin{aligned} & \\dfrac{7}{{{x}^{2}}-1}+\\dfrac{8x}{{{x}^{2}}-2x+1}=\\dfrac{25-9x}{{{x}^{3}}-{{x}^{2}}-x+1} \\\\ & \\Leftrightarrow \\dfrac{7}{\\left( x-1 \\right)\\left( x+1 \\right)}+\\dfrac{8x}{{{\\left( x-1 \\right)}^{2}}}=\\dfrac{25-9x}{{{\\left( x-1 \\right)}^{2}}\\left( x+1 \\right)} \\\\ \\end{aligned}$<br\/>Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & 7\\left( x-1 \\right)+8x\\left( x+1 \\right)=25-9x \\\\ & \\Leftrightarrow 7x-7+8x^2+8x=25-9x \\\\ & \\Leftrightarrow 8x^2+24x-32=0\\\\ & \\Leftrightarrow x^2+3x-4=0 \\\\ \\end{aligned}$<br\/>$\\Leftrightarrow \\left[ \\begin{align} & x=1 \\\\ & x=-4\\\\ \\end{align} \\right.$ (do $a+b+c=0$) <br\/>$\\Leftrightarrow x=-4$ (do $x \\ne \\pm 1)$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m l\u00e0 $x=-4$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-4.$ <\/span><\/span>"}]}],"id_ques":952},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\left( \\dfrac{x+2}{{{x}^{2}}-2x+1}-\\dfrac{x-2}{{{x}^{2}}-1} \\right).\\dfrac{x+1}{x}=2$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 ","select":["A. $S=\\left\\{-2;4 \\right\\}$ ","B. $S=\\left\\{2;-4 \\right\\}$ ","C. $S=\\left\\{1\\pm \\sqrt{3} \\right\\}$ ","D. $S=\\left\\{-1\\pm \\sqrt{3} \\right\\}$ "],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x \\ne \\{ \\pm 1; 0 \\}$ <br\/>$\\begin{aligned} & \\left( \\dfrac{x+2}{{{x}^{2}}-2x+1}-\\dfrac{x-2}{{{x}^{2}}-1} \\right).\\dfrac{x+1}{x}=2 \\\\ & \\Leftrightarrow \\left[ \\dfrac{x+2}{{{\\left( x-1 \\right)}^{2}}}-\\dfrac{x-2}{\\left( x-1 \\right)\\left( x+1 \\right)} \\right].\\dfrac{x+1}{x}=2 \\\\ \\end{aligned}$<br\/>Quy \u0111\u1ed3ng v\u00e0 r\u00fat g\u1ecdn v\u1ebf tr\u00e1i ph\u01b0\u01a1ng tr\u00ecnh, ta c\u00f3:<br\/> $\\begin{aligned} & \\dfrac{\\left( x+2 \\right)\\left( x+1 \\right)-\\left( x-2 \\right)\\left( x-1 \\right)}{{{\\left( x-1 \\right)}^{2}}\\left( x+1 \\right)}.\\dfrac{x+1}{x}=2 \\\\ & \\Leftrightarrow \\dfrac{{{x}^{2}}+3x+2-\\left( {{x}^{2}}-3x+2 \\right)}{{{\\left( x-1 \\right)}^{2}}\\left( x+1 \\right)}.\\dfrac{x+1}{x}=2 \\\\ & \\Leftrightarrow \\dfrac{6x}{{{\\left( x-1 \\right)}^{2}}\\left( x+1 \\right)}.\\dfrac{x+1}{x}=2 \\\\ & \\Leftrightarrow \\dfrac{6}{{{\\left( x-1 \\right)}^{2}}}=2 \\\\ & \\Leftrightarrow {{\\left( x-1 \\right)}^{2}}=3 \\\\ & \\Leftrightarrow x-1=\\pm \\sqrt{3} \\\\ & \\Leftrightarrow x=1\\pm \\sqrt{3} \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 $S=\\left\\{1\\pm \\sqrt{3} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":953},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\left( \\dfrac{\\sqrt{x}}{\\sqrt{x}-1}-\\dfrac{1}{x-\\sqrt{x}} \\right):\\left( \\dfrac{1}{1+\\sqrt{x}}+\\dfrac{2}{x-1} \\right)=2\\sqrt{x}-2$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 s\u1ed1 nghi\u1ec7m l\u00e0 _input_","hint":"R\u00fat g\u1ecdn v\u1ebf tr\u00e1i ph\u01b0\u01a1ng tr\u00ecnh","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x>0; x \\ne 1$ <br\/>$\\begin{aligned} & \\left( \\dfrac{\\sqrt{x}}{\\sqrt{x}-1}-\\dfrac{1}{x-\\sqrt{x}} \\right):\\left( \\dfrac{1}{1+\\sqrt{x}}+\\dfrac{2}{x-1} \\right)=2\\sqrt{x}-2 \\\\ & \\Leftrightarrow \\dfrac{x-1}{\\sqrt{x}\\left( \\sqrt{x}-1 \\right)}:\\dfrac{\\sqrt{x}+1}{\\left( \\sqrt{x}-1 \\right)\\left( \\sqrt{x}+1 \\right)}=2\\sqrt{x}-2 \\\\ & \\Leftrightarrow \\dfrac{x-1}{\\sqrt{x}\\left( \\sqrt{x}-1 \\right)}.\\dfrac{\\left( \\sqrt{x}-1 \\right)\\left( \\sqrt{x}+1 \\right)}{\\sqrt{x}+1}=2\\sqrt{x}-2 \\\\ & \\Leftrightarrow \\dfrac{x-1}{\\sqrt{x}}=2\\sqrt{x}-2 \\\\ \\end{aligned}$<br\/>Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & x-1=\\left( 2\\sqrt{x}-2 \\right)\\sqrt{x} \\\\ & \\Leftrightarrow 2x-2\\sqrt{x}=x-1 \\\\ & \\Leftrightarrow x-2\\sqrt{x}+1=0 \\\\ & \\Leftrightarrow {{\\left( \\sqrt{x}-1 \\right)}^{2}}=0\\\\ \\end{aligned}$<br\/>$\\Leftrightarrow \\sqrt{x}-1=0\\Leftrightarrow \\sqrt{x}=1\\Leftrightarrow x=1 \\,(\\text{lo\u1ea1i v\u00ec kh\u00f4ng th\u1ecfa m\u00e3n}\\, x \\ne 1)$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $0.$ <\/span><\/span>"}]}],"id_ques":954},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["5"],["7"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $4{{\\left( x-6 \\right)}^{8}}+{{\\left( x-6 \\right)}^{4}}-5=0$ (1) <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{$_input_;_input_$\\}$","hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 ${{\\left( x-6 \\right)}^{4}}=t,$ $t \\ge 0 $ . ","explain":"<span class='basic_left'>\u0110\u1eb7t ${{\\left( x-6 \\right)}^{4}}=t,t\\ge 0.$ Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: $4{{t}^{2}}+t-5=0$ (2)<br\/>V\u00ec $a+b+c=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 hai nghi\u1ec7m l\u00e0: <br\/>${{t}_{1}}=1$ (th\u1ecfa m\u00e3n); ${{t}_{2}}=-5$ (lo\u1ea1i v\u00ec kh\u00f4ng th\u1ecfa m\u00e3n $t\\ge 0$)<br\/>V\u1edbi $t={{t}_{1}}=1,$ ta c\u00f3 ${{\\left( x-6 \\right)}^{4}}=1\\Leftrightarrow x-6=\\pm 1\\Leftrightarrow \\left[ \\begin{align} & x=7 \\\\ & x=5 \\\\ \\end{align} \\right.$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ 5;7 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $5;$ $7.$ <\/span><\/span>"}]}],"id_ques":955},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\sqrt{2}{{x}^{3}}+3{{x}^{2}}-2=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0","select":["A. $S=\\left\\{ \\sqrt{2};\\dfrac{\\sqrt{2}}{2} \\right\\}$","B. $S=\\left\\{ -\\sqrt{2};\\dfrac{\\sqrt{2}}{2} \\right\\}$","C. $S=\\left\\{ -\\sqrt{2};-\\dfrac{\\sqrt{2}}{2} \\right\\}$ ","D. $S=\\left\\{ -\\dfrac{\\sqrt{2}}{2};\\dfrac{\\sqrt{2}}{2} \\right\\}$"],"hint":"Nh\u00e2n th\u00eam $2$ v\u00e0o hai v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh sau \u0111\u00f3 \u0111\u1eb7t \u1ea9n ph\u1ee5 $y=x\\sqrt{2}$","explain":"<span class='basic_left'>$\\sqrt{2}{{x}^{3}}+3{{x}^{2}}-2=0\\Leftrightarrow 2\\sqrt{2}{{x}^{3}}+2.3{{x}^{2}}-4=0$<br\/>\u0110\u1eb7t $y=x\\sqrt{2}$ , ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: ${{y}^{3}}+3{{y}^{2}}-4=0$ (*)<br\/> V\u00ec t\u1ed5ng c\u00e1c h\u1ec7 s\u1ed1 b\u1eb1ng $0$ n\u00ean (*) c\u00f3 m\u1ed9t nghi\u1ec7m b\u1eb1ng $1$.<br\/> Khi \u0111\u00f3, b\u1eb1ng ph\u00e9p chia \u0111a th\u1ee9c v\u1ebf tr\u00e1i cho \u0111a th\u1ee9c $y-1,$ ta c\u00f3: <br\/>$\\begin{aligned} & (y-1)(y^2+4y+4)=0 \\\\ & \\Leftrightarrow \\left( y-1 \\right){{\\left( y+2 \\right)}^{2}}=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & y=1 \\\\ & y=-2 \\\\ \\end{aligned} \\right.\\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x\\sqrt{2}=1 \\\\ & x\\sqrt{2}=-2 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{\\sqrt{2}}{2} \\\\ & x=-\\sqrt{2} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Suy ra t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ -\\sqrt{2};\\dfrac{\\sqrt{2}}{2} \\right\\}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":956},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{4}}+4{{x}^{2}}+4=16{{x}^{2}}-8x+1$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0","select":["A. $S=\\left\\{ 1;3;2+\\sqrt{3};2-\\sqrt{3} \\right\\}$","B. $S=\\left\\{ -1;-3;-2+\\sqrt{3};-2-\\sqrt{3} \\right\\}$","C. $S=\\left\\{ -1;-3;2+\\sqrt{3};2-\\sqrt{3} \\right\\}$ ","D. $S=\\left\\{ 1;3;-2+\\sqrt{3};-2-\\sqrt{3} \\right\\}$"],"hint":"X\u00e9t xem hai v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec1u l\u00e0 d\u1ea1ng khai tri\u1ec3n c\u1ee7a h\u1eb1ng \u0111\u1eb3ng th\u1ee9c n\u00e0o?","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$\\begin{aligned} & {{x}^{4}}+4{{x}^{2}}+4=16{{x}^{2}}-8x+1 \\\\ & \\Leftrightarrow {{\\left( {{x}^{2}}+2 \\right)}^{2}}={{\\left( 4x-1 \\right)}^{2}} \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}+2=4x-1 \\\\ & {{x}^{2}}+2=-\\left( 4x-1 \\right) \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}-4x+3=0\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{x}^{2}}+4x+1=0\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 $a+b+c=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 $1$ v\u00e0 $3$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 $\\Delta '=4-1=3$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $x=-2\\pm \\sqrt{3}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ 1;3;-2+\\sqrt{3};-2-\\sqrt{3} \\right\\}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":957},{"time":24,"part":[{"title":"Gi\u1ea3 s\u1eed ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{4}}+b{{x}^{2}}+c=0$ v\u1edbi $a\\ne 0$ c\u00f3 4 nghi\u1ec7m ph\u00e2n bi\u1ec7t. Khi \u0111\u00f3:","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["t","f"]],"list":[{"point":10,"image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["T\u1ed5ng c\u00e1c nghi\u1ec7m b\u1eb1ng $0$","T\u00edch c\u00e1c nghi\u1ec7m b\u1eb1ng ${{\\left( \\dfrac{c}{a} \\right)}^{2}}$ "],"hint":"","explain":["<span class='basic_left'> \u0110\u00fang v\u00ec \u0111\u1eb7t $t=x^2, $ ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: $at^2+bt+c=0$ (*)<br\/>N\u1ebfu ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng c\u00f3 4 nghi\u1ec7m th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t $t=\\dfrac{-b\\pm \\sqrt{\\Delta} }{2a}$<br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{4}}+b{{x}^{2}}+c=0$ c\u00f3 4 nghi\u1ec7m l\u00e0:<br\/> ${{x}_{1,2}}=\\pm \\sqrt{\\dfrac{-b+\\sqrt{\\Delta }}{2a}};{{x}_{3,4}}=\\pm \\sqrt{\\dfrac{-b-\\sqrt{\\Delta }}{2a}}$<br\/>Suy ra ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=0$ (c\u00f3 hai c\u1eb7p nghi\u1ec7m \u0111\u1ed1i nhau)<\/span>","<br\/><span class='basic_left'>Sai v\u00ec<br\/>$\\begin{aligned} & {{x}_{1}}.{{x}_{2}}.{{x}_{3}}.{{x}_{4}}=\\left( \\pm \\sqrt{\\dfrac{-b+\\sqrt{\\Delta }}{2a}} \\right).\\left( \\pm \\sqrt{\\dfrac{-b-\\sqrt{\\Delta }}{2a}} \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\dfrac{-b+\\sqrt{\\Delta }}{2a}.\\dfrac{-b-\\sqrt{\\Delta }}{2a}=\\dfrac{{{b}^{2}}-\\Delta }{4{{a}^{2}}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\dfrac{4ac}{4{{a}^{2}}}=\\dfrac{c}{a} \\\\ \\end{aligned}$<\/span>"]}]}],"id_ques":958},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{4}}-2\\left( m+1 \\right){{x}^{2}}+{{m}^{2}}=0$ (1).<br\/>T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 4 nghi\u1ec7m ph\u00e2n bi\u1ec7t. ","select":["A. $m>-1$ v\u00e0 $m\\ne 0$","B. $m<-\\dfrac{1}{2}$","C. $m>-\\dfrac{1}{2}$ v\u00e0 $m\\ne 0$"],"hint":"\u0110\u1eb7t $t=x^2$. Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 4 nghi\u1ec7m ph\u00e2n bi\u1ec7t $\\Leftrightarrow $ ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai \u1ea9n $t$ c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t","explain":"<span class='basic_left'>\u0110\u1eb7t $t={{x}^{2}},t\\ge 0$. Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh:<br\/> ${{t}^{2}}-2\\left( m+1 \\right)t+{{m}^{2}}=0$ <br\/>$\\Delta '={{\\left( m+1 \\right)}^{2}}-{{m}^{2}}=2m+1;S=\\dfrac{-b}{a}=2\\left( m+1 \\right),P=\\dfrac{c}{a}={{m}^{2}}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 4 nghi\u1ec7m ph\u00e2n bi\u1ec7t $\\Leftrightarrow $ ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t<br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & \\Delta '>0 \\\\ & S>0 \\\\ & P>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2m+1>0 \\\\ & 2\\left( m+1 \\right)>0 \\\\ & {{m}^{2}}>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m>-\\dfrac{1}{2} \\\\ & m>-1 \\\\ & m\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m>-\\dfrac{1}{2} \\\\ & m\\ne 0 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":3}]}],"id_ques":959},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{4}}-2\\left( m+1 \\right){{x}^{2}}+{{m}^{2}}=0$ (1).<br\/>T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. ","select":["A. $m>-1$ v\u00e0 $m\\ne 0$","B. $m<-\\dfrac{1}{2}$","C. $m>-\\dfrac{1}{2}$ v\u00e0 $m\\ne 0$"],"hint":"\u0110\u1eb7t $t=x^2$. Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00f4 nghi\u1ec7m $\\Leftrightarrow $ ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai \u1ea9n $t$ v\u00f4 nghi\u1ec7m ho\u1eb7c c\u00f3 hai nghi\u1ec7m \u00e2m","explain":"<span class='basic_left'>\u0110\u1eb7t $t={{x}^{2}},t\\ge 0$. Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh:<br\/> ${{t}^{2}}-2\\left( m+1 \\right)t+{{m}^{2}}=0$ <br\/>$\\Delta '={{\\left( m+1 \\right)}^{2}}-{{m}^{2}}=2m+1;S=\\dfrac{-b}{a}=2\\left( m+1 \\right),P=\\dfrac{c}{a}={{m}^{2}}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00f4 nghi\u1ec7m $\\Leftrightarrow $ ph\u01b0\u01a1ng tr\u00ecnh (2) v\u00f4 nghi\u1ec7m ho\u1eb7c c\u00f3 hai nghi\u1ec7m \u00e2m<br\/>+ Tr\u01b0\u1eddng h\u1ee3p 1: Ph\u01b0\u01a1ng tr\u00ecnh (2) v\u00f4 nghi\u1ec7m $\\Leftrightarrow \\Delta '<0\\Leftrightarrow 2m+1<0\\Leftrightarrow m<-\\dfrac{1}{2}$ <br\/>+ Tr\u01b0\u1eddng h\u1ee3p 2: (2) c\u00f3 hai nghi\u1ec7m \u00e2m $\\Leftrightarrow \\left\\{ \\begin{aligned} & \\Delta '\\ge 0 \\\\ & S<0 \\\\ & P>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2m+1\\ge 0 \\\\ & 2\\left( m+1 \\right)<0 \\\\ & {{m}^{2}}>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ge -\\dfrac{1}{2} \\\\ & m<-1 \\\\ & m\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow m\\in \\varnothing $ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00f4 nghi\u1ec7m khi $m<-\\dfrac{1}{2}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":3}]}],"id_ques":960}],"lesson":{"save":0,"level":2}}