{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng trong b\u1ea3ng","title_trans":"","temp":"fill_the_blank","correct":[[["10"],["16"],["-7"],["10"],["2"],["-3"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Bi\u1ebft c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau c\u00f3 hai nghi\u1ec7m ${{x}_{1}}$ v\u00e0 ${{x}_{2}}$. T\u00ednh t\u1ed5ng v\u00e0 t\u00edch hai nghi\u1ec7m r\u1ed3i \u0111i\u1ec1n v\u00e0o b\u1ea3ng sau:<\/span><br\/><table><tr><th>Ph\u01b0\u01a1ng tr\u00ecnh<br><\/th><th>${{x}_{1}}+{{x}_{2}}$ <br><\/th><th>${{x}_{1}}.{{x}_{2}}$ <br><\/th><\/tr><tr><td>${{x}^{2}}-10x+16=0$ <\/td><td>_input_<\/td><td>_input_<\/td><\/tr><tr><td>${{x}^{2}}+7x+10=0$ <\/td><td>_input_<\/td><td>_input_<\/td><\/tr><tr><td>$2{{x}^{2}}-4x-6=0$ <\/td><td>_input_<\/td><td>_input_<\/td><\/tr><\/table>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi \u2013 \u00e9t:<br\/> N\u1ebfu ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ $\\left( a\\ne 0 \\right)$ c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ th\u00ec $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=-\\dfrac{b}{a} \\\\ & {{x}_{1}}.{{x}_{2}}=\\dfrac{c}{a} \\\\ \\end{align} \\right.$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-10x+16=0$ c\u00f3 ${{x}_{1}}+{{x}_{2}}=10;{{x}_{1}}.{{x}_{2}}=16$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+7x+10=0$ c\u00f3 ${{x}_{1}}+{{x}_{2}}=-7;{{x}_{1}}.{{x}_{2}}=10$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{2}}-4x-6=0$ c\u00f3 ${{x}_{1}}+{{x}_{2}}=-\\dfrac{-4}{2}=2;{{x}_{1}}.{{x}_{2}}=\\dfrac{-6}{2}=-3$<br\/>Ta c\u00f3 b\u1ea3ng k\u1ebft qu\u1ea3:<br\/><table><tr><th>Ph\u01b0\u01a1ng tr\u00ecnh<br><\/th><th>${{x}_{1}}+{{x}_{2}}$ <br><\/th><th>${{x}_{1}}.{{x}_{2}}$ <br><\/th><\/tr><tr><td>${{x}^{2}}-10x+16=0$ <\/td><td>$10$<\/td><td>$16$<\/td><\/tr><tr><td>${{x}^{2}}+7x+10=0$ <\/td><td>$-7$<\/td><td>$10$<\/td><\/tr><tr><td>$2{{x}^{2}}-4x-6=0$ <\/td><td>$2$<\/td><td>$-3$<\/td><\/tr><\/table><\/span><\/span>"}]}],"id_ques":871},{"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":"Ph\u01b0\u01a1ng tr\u00ecnh $4{{x}^{2}}-4x+3=0$ c\u00f3 t\u1ed5ng c\u00e1c nghi\u1ec7m b\u1eb1ng $1$ v\u00e0 t\u00edch c\u00e1c nghi\u1ec7m b\u1eb1ng $\\dfrac{3}{4},$ <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?","select":["A. \u0110\u00fang ","B. Sai"],"hint":"Ki\u1ec3m tra xem ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m hay kh\u00f4ng.","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh $4{{x}^{2}}-4x+3=0$ c\u00f3 $\\Delta '=4-12=-8<0$ n\u00ean v\u00f4 nghi\u1ec7m<br\/>V\u00ec v\u1eady kh\u00f4ng th\u1ec3 n\u00f3i \u0111\u1ebfn t\u1ed5ng v\u00e0 t\u00edch c\u00e1c nghi\u1ec7m<br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 sai<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span><br\/>\u0110\u1ec3 t\u00ednh \u0111\u01b0\u1ee3c t\u1ed5ng v\u00e0 t\u00edch hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh theo h\u1ec7 th\u1ee9c Vi - \u00e9t, ta c\u1ea7n ki\u1ec3m tra ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u00e0 c\u00f3 nghi\u1ec7m hay kh\u00f4ng. <\/span>","column":2}]}],"id_ques":872},{"time":24,"part":[{"title":"Cho ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{2}}+9x-5=0$ ","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["f","t","f"]],"list":[{"point":5,"image":"img\/1.png","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["Ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m c\u00f9ng d\u1ea5u","$-5$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh","$-\\dfrac{1}{2}$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh."],"hint":"","explain":["Sai v\u00ec ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 $a.c=-10<0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u.","<br\/>\u0110\u00fang v\u00ec v\u1edbi $x=-5,$ th\u00ec $2{{x}^{2}}+9x-5=2.{{\\left( -5 \\right)}^{2}}+9.\\left( -5 \\right)-5=0$ n\u00ean $-5$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho.","<br\/>Sai: Theo kh\u1eb3ng \u0111\u1ecbnh 2 th\u00ec $x=-5$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh n\u00ean theo h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: ${{x}_{1}}.{{x}_{2}}=-\\dfrac{5}{2}\\Rightarrow {{x}_{2}}=-\\dfrac{5}{2}:{{x}_{1}}=-\\dfrac{5}{2}:\\left( -5 \\right)=\\dfrac{1}{2}$ "]}]}],"id_ques":873},{"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":"Hai s\u1ed1 $5$ v\u00e0 $-8$ l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh","select":["A. ${{x}^{2}}-3x+40=0$ ","B. ${{x}^{2}}-3x-40=0$","C. ${{x}^{2}}+3x-40=0$ ","D. ${{x}^{2}}+13x-40=0$ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>B\u01b0\u1edbc 1: T\u00ednh t\u1ed5ng hai nghi\u1ec7m $S={{x}_{1}}+{{x}_{2}}$ v\u00e0 t\u00edch hai nghi\u1ec7m $P={{x}_{1}}{{x}_{2}}$ <br\/>B\u01b0\u1edbc 2: Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ l\u00e0 ${{x}^{2}}-Sx+P=0$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3 t\u1ed5ng $S=5+\\left( -8 \\right)=-3$ v\u00e0 t\u00edch $P=5.\\left( -8 \\right)=-40$ <br\/>V\u1eady hai s\u1ed1 $5$ v\u00e0 $-8$ l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+3x-40=0$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":874},{"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"],["-4"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 hai nghi\u1ec7m b\u1eb1ng $1-\\sqrt{5}$ v\u00e0 $1+\\sqrt{5}$ <br\/><b>\u0110\u00e1p s\u1ed1: <\/b> Ph\u01b0\u01a1ng tr\u00ecnh l\u1eadp \u0111\u01b0\u1ee3c l\u00e0: $X^2+$_input_$X+$_input_$=0$ ","hint":"T\u00ednh t\u1ed5ng v\u00e0 t\u00edch hai s\u1ed1 r\u1ed3i l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh","explain":"<span class='basic_left'>Ta c\u00f3 t\u1ed5ng $S=\\left( 1-\\sqrt{5} \\right)+\\left( 1+\\sqrt{5} \\right)=2$ v\u00e0 t\u00edch $P=\\left( 1-\\sqrt{5} \\right)\\left( 1+\\sqrt{5} \\right)=1-5=-4$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 hai nghi\u1ec7m b\u1eb1ng $1-\\sqrt{5}$ v\u00e0 $1+\\sqrt{5}$ l\u00e0 ${{X}^{2}}-2X-4=0$ <br\/> <span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $-2$ v\u00e0 $-4.$<\/span><\/span>"}]}],"id_ques":875},{"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":"T\u00ecm hai s\u1ed1 $a$ v\u00e0 $b$ bi\u1ebft $a+b=10;$ $a.b=21$ ","select":["A. $a=7;b=3$","B. $a=3;b=7$","C. $a=-3;b=13$","D. C\u1ea3 A v\u00e0 B"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u00e0i to\u00e1n: T\u00ecm hai s\u1ed1 bi\u1ebft t\u1ed5ng $S$ v\u00e0 t\u00edch $P$<br\/>Hai s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai $X^2-SX+P=0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Hai s\u1ed1 $a, b$ c\u1ea7n t\u00ecm l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai $X^2-10X+21=0$<br\/>$\\Delta '=25-21=4$<br\/> Suy ra ${{X}_{1}}=5+2=7;{{X}_{2}}=5-2=3$ <br\/>V\u1eady $a=7;b=3$ ho\u1eb7c $a=3;b=7$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t: <\/span>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 t\u1ed3n t\u1ea1i hai s\u1ed1 $a, b$ l\u00e0 ${{S}^{2}}\\ge 4P$ <\/span>","column":2}]}],"id_ques":876},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n d\u01b0\u1edbi d\u1ea1ng s\u1ed1 nguy\u00ean ho\u1eb7c s\u1ed1 th\u1eadp ph\u00e2n","temp":"fill_the_blank_random","correct":[[["3"],["-2,5"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>T\u00ecm hai s\u1ed1 bi\u1ebft t\u1ed5ng c\u1ee7a ch\u00fang l\u00e0 $\\dfrac{1}{2}$ v\u00e0 t\u00edch c\u1ee7a ch\u00fang l\u00e0 $-\\dfrac{15}{2}$ <br\/>Hai s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 _input_ v\u00e0 _input_<\/span>","hint":"","explain":"<span class='basic_left'>G\u1ecdi hai s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 $u$ v\u00e0 $v$. Theo \u0111\u1ec1 ra, ta c\u00f3: <br\/>$u+v=\\dfrac{1}{2}$ v\u00e0 $u.v=-\\dfrac{15}{2}$<br\/>Khi \u0111\u00f3, hai s\u1ed1 $u,$ $v$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai<br\/>$\\begin{align} & {{X}^{2}}-\\dfrac{1}{2}X-\\dfrac{15}{2}=0 \\\\ & \\Leftrightarrow 2{{X}^{2}}-X-15=0 \\\\ \\end{align}$ <br\/>$\\Delta =1-4.2\\left( -15 \\right)=121$ <br\/>Suy ra ${{X}_{1}}=\\dfrac{1+11}{4}=3;{{X}_{2}}=\\dfrac{1-11}{4}=-\\dfrac{5}{2}$ <br\/>V\u1eady $u=3;v=-\\dfrac{5}{2}$ ho\u1eb7c $u=-\\dfrac{5}{2};v=3$ hay hai s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 $3$ v\u00e0 $-2,5$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3$ v\u00e0 $-2,5$<\/span><\/span>"}]}],"id_ques":877},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{5}{3}$","B. $\\dfrac{1}{3}$","C. $\\dfrac{1}{4}$"],"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+5x-3=0$. G\u1ecdi ${{x}_{1}},{{x}_{2}}$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. <br\/>Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $A=\\dfrac{1}{{{x}_{1}}}+\\dfrac{1}{{{x}_{2}}}$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b>$A=$?<br\/>(\u0110i\u1ec1n \u0111\u00e1p \u00e1n d\u01b0\u1edbi d\u1ea1ng ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n)<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t \u0111\u1ec3 t\u00ednh t\u1ed5ng $S$ v\u00e0 t\u00edch $P$ c\u1ee7a hai nghi\u1ec7m.<br\/>B\u01b0\u1edbc 2: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $A$ v\u1ec1 bi\u1ec3u th\u1ee9c ch\u1ec9 ch\u1ee9a $S$ v\u00e0 $P$<br\/>B\u01b0\u1edbc 3: Thay gi\u00e1 tr\u1ecb c\u1ee7a $S$ v\u00e0 $P$ \u1edf b\u01b0\u1edbc 1 v\u00e0o $A$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span> <br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+5x-3=0$ c\u00f3 $ac=-3<0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m.<br\/>Theo h\u1ec7 th\u1ee9c Vi \u2013 \u00e9t, ta c\u00f3 $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=-5 \\\\ & {{x}_{1}}.{{x}_{2}}=-3 \\\\ \\end{align} \\right.$ <br\/>$A=\\dfrac{1}{{{x}_{1}}}+\\dfrac{1}{{{x}_{2}}}=\\dfrac{{{x}_{1}}+{{x}_{2}}}{{{x}_{1}}{{x}_{2}}}=\\dfrac{-5}{-3}=\\dfrac{5}{3}$<\/span>"}]}],"id_ques":878},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["31"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+5x-3=0$. G\u1ecdi ${{x}_{1}},{{x}_{2}}$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. <br\/><b>C\u00e2u 2: <\/b>Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $B=x_{1}^{2}+x_{2}^{2}$ <br\/><b> \u0110\u00e1p s\u1ed1: <\/b>$B=$_input_<\/span>","hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+5x-3=0$ c\u00f3 $ac=-3<0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m.<br\/>Theo h\u1ec7 th\u1ee9c Vi \u2013 \u00e9t, ta c\u00f3 $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=-5 \\\\ & {{x}_{1}}.{{x}_{2}}=-3 \\\\ \\end{align} \\right.$<br\/>$\\begin{align} & B=x_{1}^{2}+x_{2}^{2} \\\\ & \\,\\,\\,\\,\\,=\\left( x_{1}^{2}+2{{x}_{1}}{{x}_{2}}+x_{2}^{2} \\right)-2{{x}_{1}}{{x}_{2}} \\\\ & \\,\\,\\,\\,\\,={{\\left( {{x}_{2}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}} \\\\ & \\,\\,\\,\\,\\,={{\\left( -5 \\right)}^{2}}-2.\\left( -3 \\right) \\\\ & \\,\\,\\,\\,\\,=25+6=31 \\\\ \\end{align}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $31.$<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>Bi\u1ec3u th\u1ee9c $A$ v\u00e0 $B$ nh\u01b0 tr\u00ean l\u00e0 c\u00e1c bi\u1ec3u th\u1ee9c \u0111\u1ed1i x\u1ee9ng gi\u1eefa $x_1$ v\u00e0 $x_2$. <br\/>Ngo\u00e0i ra, c\u00e1c em c\u00f2n g\u1eb7p nhi\u1ec1u bi\u1ec3u th\u1ee9c \u0111\u1ed1i x\u1ee9ng kh\u00e1c nh\u01b0:<br\/>$x_{1}^{3}+x_{2}^{3}$ ; $\\dfrac{{{x}_{1}}}{{{x}_{2}}}+\\dfrac{{{x}_{2}}}{{{x}_{1}}};\\dfrac{x_{1}^{2}}{x_{2}^{{}}}+\\dfrac{x_{2}^{2}}{{{x}_{1}}};...$ <\/span>"}]}],"id_ques":879},{"time":24,"part":[{"time":3,"title":"Nh\u1ea9m nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai 1 \u1ea9n","title_trans":"N\u1ed1i ph\u01b0\u01a1ng tr\u00ecnh \u1edf c\u1ed9t b\u00ean tr\u00e1i v\u1edbi nghi\u1ec7m \u1edf c\u1ed9t b\u00ean ph\u1ea3i \u0111\u1ec3 \u0111\u01b0\u1ee3c \u0111\u00e1p \u00e1n \u0111\u00fang","audio":"","temp":"matching","correct":[["3","2","4","1"]],"list":[{"point":5,"image":"img\/1.png","left":["Ph\u01b0\u01a1ng tr\u00ecnh $3{{x}^{2}}-4x+1=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 ","Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+3x-4=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 ","Ph\u01b0\u01a1ng tr\u00ecnh $2012x+2000x-12=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 ","Ph\u01b0\u01a1ng tr\u00ecnh . ${{x}^{2}}+2013x+2012=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 "],"right":["a. $ {{x}_{1}}=-1;{{x}_{2}}=-2012$ ","b. ${{x}_{1}}=1;{{x}_{2}}=-4$","c. ${{x}_{1}}=1;{{x}_{2}}=\\dfrac{1}{3}$","d. ${{x}_{1}}=-1;{{x}_{2}}=\\dfrac{3}{503}$ "],"top":80,"hint":"Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai $ax^2+bx+c=0, a \\ne 0$<br\/>N\u1ebfu $a+b+c=0$ th\u00ec ${{x}_{1}}=1;{{x}_{2}}=\\dfrac{c}{a}$<br\/>N\u1ebfu $a-b+c=0$ th\u00ec ${{x}_{1}}=-1;{{x}_{2}}=-\\dfrac{c}{a}$","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh $3{{x}^{2}}-4x+1=0$ c\u00f3 $a+b+c=3-4+1=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=1;{{x}_{2}}=\\dfrac{1}{3}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+3x-4=0$ c\u00f3 $a+b+c=1+3-4=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=1;{{x}_{2}}=-4$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh $2012x+2000x-12=0$ c\u00f3 $a-b+c=2012-2000-12=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 <br\/>${{x}_{1}}=-1;{{x}_{2}}=\\dfrac{12}{2012}=\\dfrac{3}{503}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+2013x+2012=0$ c\u00f3 $a-b+c=1-2013+2012=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0<br\/> ${{x}_{1}}=-1;{{x}_{2}}=-2012$ <br\/><span class='basic_pink'>V\u1eady 1 n\u1ed1i v\u1edbi c, 2 n\u1ed1i v\u1edbi b, 3 n\u1ed1i v\u1edbi d v\u00e0 4 n\u1ed1i v\u1edbi a.<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>+ \u0110\u1ec3 s\u1eed d\u1ee5ng \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng ph\u00e1p nh\u1ea9m nghi\u1ec7m th\u00ec ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho ph\u1ea3i l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai.<br\/>+ Khi t\u00ednh nh\u1ea9m nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng b\u1eadc hai, tr\u01b0\u1edbc h\u1ebft ta th\u01b0\u1eddng x\u00e9t xem ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00f3 c\u00f3 nghi\u1ec7m l\u00e0 $1$ hay $-1$ hay kh\u00f4ng. Trong th\u1ef1c h\u00e0nh, ta th\u01b0\u1eddng t\u00ednh $a+c$ tr\u01b0\u1edbc. N\u1ebfu $a+c=b$ th\u00ec $-1$ l\u00e0 m\u1ed9t nghi\u1ec7m; $a+c=-b$ th\u00ec $1$ l\u00e0 m\u1ed9t nghi\u1ec7m.<\/span>"}]}],"id_ques":880},{"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":"Ph\u01b0\u01a1ng tr\u00ecnh $-5\\sqrt{2}{{x}^{2}}+\\sqrt{2}\\left( \\sqrt{2}+5 \\right)x-2=0$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $-\\dfrac{\\sqrt{2}}{5},$ <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?","select":["A. \u0110\u00fang ","B. Sai"],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh $-5\\sqrt{2}{{x}^{2}}+\\sqrt{2}\\left( \\sqrt{2}+5 \\right)x-2=0$ c\u00f3 $a+b+c=-5\\sqrt{2}+\\sqrt{2}\\left( \\sqrt{2}+5 \\right)-2=0$<br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m l\u00e0 ${{x}_{1}}=1;{{x}_{2}}=\\dfrac{2}{5\\sqrt{2}}=\\dfrac{\\sqrt{2}}{5}$ <br\/>Suy ra $x=-\\dfrac{\\sqrt{2}}{5}$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span <\/span>","column":2}]}],"id_ques":881},{"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":[[["3"],["5"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"T\u00ednh nh\u1ea9m nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-\\left( \\sqrt{3}+\\sqrt{5} \\right)x+\\sqrt{15}=0$ <br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $ x_1=\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ v\u00e0 $x_2=\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$","hint":"Nh\u1ea9m: ${{x}_{1}}+{{x}_{2}}=m+n;{{x}_{1}}.{{x}_{2}}=mn$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m ${{x}_{1}}=m;{{x}_{2}}=n$ ","explain":"Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ${{x}_{1}}+{{x}_{2}}=\\sqrt{3}+\\sqrt{5};{{x}_{1}}.{{x}_{2}}=\\sqrt{15}=\\sqrt{3}.\\sqrt{5}$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 ${{x}_{1}}=\\sqrt{3};{{x}_{2}}=\\sqrt{5}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3$ v\u00e0 $5.$<\/span>"}]}],"id_ques":882},{"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":[[["-7"],["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Bi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh $x^2+mx-35=0$ c\u00f3 1 nghi\u1ec7m ${{x}_{1}}=5$. T\u00ecm nghi\u1ec7m ${{x}_{2}}$ c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh r\u1ed3i t\u1eeb \u0111\u00f3 t\u00ecm $m$<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $x_2=$_input_; $m=$_input_","hint":"\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t \u0111\u1ec3 t\u00ecm nghi\u1ec7m $x_2$ c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. ","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh $x^2+mx-35=0$ c\u00f3 hai nghi\u1ec7m $x_1;x_2$ n\u00ean theo h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3:<br\/> ${{x}_{1}}.{{x}_{2}}=-35\\Leftrightarrow {{x}_{2}}=-35:{{x}_{1}}=-35:5=-7$ <br\/>Khi \u0111\u00f3 ${{x}_{1}}+{{x}_{2}}=-m\\Leftrightarrow 5+\\left( -7 \\right)=-m\\Leftrightarrow m=2$ <br\/>V\u1eady ${{x}_{2}}=-7;m=2$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-7$ v\u00e0 $2$<\/span>"}]}],"id_ques":883},{"time":24,"part":[{"time":3,"title":"X\u00e9t d\u1ea5u c\u00e1c nghi\u1ec7m c\u1ee7a c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh","title_trans":"N\u1ed1i ph\u01b0\u01a1ng tr\u00ecnh \u1edf c\u1ed9t b\u00ean tr\u00e1i v\u1edbi k\u1ebft lu\u1eadn \u1edf c\u1ed9t b\u00ean ph\u1ea3i \u0111\u1ec3 \u0111\u01b0\u1ee3c \u0111\u00e1p \u00e1n \u0111\u00fang","audio":"","temp":"matching","correct":[["3","1","2"]],"list":[{"point":5,"image":"img\/1.png","left":[" $2{{x}^{2}}-7x+3=0$ ","$5{{x}^{2}}+3x+1=0$ ","$3x^2+7x-1=0$"],"right":["a. ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t ","b. ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u","c. ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t"],"top":80,"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/>Cho ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai $ax^2+bx+c=0$ c\u00f3 nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$.<br\/> \u0110\u1eb7t $S={{x}_{1}}+{{x}_{2}}=-\\dfrac{b}{a};P={{x}_{1}}{{x}_{2}}=\\dfrac{c}{a}$ <br\/>Ta c\u00f3 \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 m\u1ed9t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai:<br\/>+ C\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u l\u00e0: $P<0$ hay $ac<0$<br\/>+ C\u00f3 hai nghi\u1ec7m c\u00f9ng d\u1ea5u l\u00e0 $\\left\\{ \\begin{align} & \\Delta \\ge 0 \\\\ & P>0 \\\\ \\end{align} \\right.$ <br\/>+ C\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng l\u00e0 $\\left\\{ \\begin{align} & \\Delta \\ge 0 \\\\ & P>0 \\\\ & S>0 \\\\ \\end{align} \\right.$<br\/>+ C\u00f3 hai nghi\u1ec7m \u00e2m l\u00e0 $\\left\\{ \\begin{align} & \\Delta \\ge 0 \\\\ & P>0 \\\\ & S<0 \\\\ \\end{align} \\right.$<br\/>N\u1ebfu $\\Delta >0$ th\u00ec ta th\u00eam \u201cph\u00e2n bi\u1ec7t\u201d.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{2}}-7x+3=0$ c\u00f3 $\\Delta =25>0;S=\\dfrac{7}{3}>0;P=\\dfrac{3}{2}>0.$ <br\/> Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t<br\/>Ph\u01b0\u01a1ng tr\u00ecnh $5{{x}^{2}}+6x+1=0$c\u00f3 $\\Delta' =4>0;S=-\\dfrac{6}{5}<0;P=\\dfrac{1}{5}>0.$ <br\/> Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t<br\/>Ph\u01b0\u01a1ng tr\u00ecnh $3x^2+7x-1=0$ c\u00f3 $ac=-3<0$ <br\/> Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u<br\/><span class='basic_pink'>V\u1eady 1 n\u1ed1i v\u1edbi c, 2 n\u1ed1i v\u1edbi a v\u00e0 3 n\u1ed1i v\u1edbi b<\/span><\/span>"}]}],"id_ques":884},{"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":"T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-2mx+{{\\left( m-1 \\right)}^{2}}=0$ c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng","select":["A. $m > \\dfrac{1}{2}$ v\u00e0 $m\\ne 0$ ","B. $m \\ge \\dfrac{1}{2}$ v\u00e0 $m\\ne 1$","C. $m \\ge -\\dfrac{1}{2}$ v\u00e0 $m\\ne 1$","D. $m \\ne 1$ v\u00e0 $m\\ne 0$"],"hint":"Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng $\\Leftrightarrow \\Delta '\\ge 0; P>0$ v\u00e0 $S>0$","explain":"<span class='basic_left'>Ta c\u00f3 <br\/>$\\begin{aligned} & \\Delta '={{m}^{2}}-{{\\left( m-1 \\right)}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,={{m}^{2}}-\\left( {{m}^{2}}-2m+1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,=2m-1 \\\\ \\end{aligned}$<br\/> Ta c\u00f3 $S=2m; P=(m-1)^2$<br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng th\u00ec $\\left\\{ \\begin{aligned} & \\Delta ' \\ge 0 \\\\ & P>0 \\\\ & S>0 \\\\ \\end{aligned} \\right.$<br\/> $\\Leftrightarrow \\left\\{ \\begin{aligned} & 2m-1\\ge 0 \\\\ & {{\\left( m-1 \\right)}^{2}}>0 \\\\ & 2m>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ge \\dfrac{1}{2} \\\\ & m\\ne 1 \\\\ & m > 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ge \\dfrac{1}{2} \\\\ & m\\ne 1 \\\\ \\end{aligned} \\right.$<br\/> V\u1eady $m\\ge \\dfrac{1}{2}$ v\u00e0 $m\\ne 1$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":885},{"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":"V\u1edbi m\u1ecdi gi\u00e1 tr\u1ecb $m$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{2}}-2\\left( m+1 \\right)x+m=0$ lu\u00f4n c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t, <b>\u0111\u00fang<\/b> hay <b> sai <\/b>?","select":["A. \u0110\u00fang ","B. Sai"],"hint":"Ki\u1ec3m tra \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t: $\\Delta' > 0; P>0$ v\u00e0 $S<0$ ","explain":"<span class='basic_left'>Ta c\u00f3 <br\/>$\\begin{aligned} & \\Delta '={{\\left( m+1 \\right)}^{2}}-2m \\\\ & \\,\\,\\,\\,\\,\\,={{m}^{2}}+2m+1-2m \\\\ & \\,\\,\\,\\,\\,\\,={{m}^{2}}+1>0 \\\\ \\end{aligned}$<br\/> Suy ra ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/>Ta c\u00f3 $S=m+1;$ $P=\\dfrac{m}{2}$<br\/> \u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t th\u00ec $\\left\\{ \\begin{aligned} & P>0 \\\\ & S<0 \\\\ \\end{aligned} \\right.$<br\/> $\\Leftrightarrow \\left\\{ \\begin{aligned} & \\dfrac{m}{2}>0 \\\\ & m+1<0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m>0 \\\\ & m<-1 \\\\ \\end{aligned} \\right.$ (kh\u00f4ng t\u1ed3n t\u1ea1i)<br\/>V\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t.<br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh b\u00e0i to\u00e1n l\u00e0 sai.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":886},{"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":[[["15"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-8x+m=0.$ T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ th\u1ecfa m\u00e3n ${{x}_{1}}-{{x}_{2}}=2$ <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m=$_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 \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m<br\/>B\u01b0\u1edbc 2: T\u1eeb h\u1ec7 th\u1ee9c Vi-\u00e9t v\u00e0 h\u1ec7 th\u1ee9c \u0111\u00e3 cho, ta gi\u1ea3i h\u1ec7 \u0111\u1ed1i v\u1edbi nghi\u1ec7m $x_1;x_2$ r\u1ed3i thay v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh c\u00f2n l\u1ea1i \u0111\u1ec3 t\u00ecm $m.$<br\/>B\u01b0\u1edbc 3: Ki\u1ec3m tra $m$ c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n kh\u00f4ng r\u1ed3i k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-8x+m=0$ c\u00f3 $\\Delta '=16-m$<br\/> Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $\\Leftrightarrow \\Delta '\\ge 0\\Leftrightarrow 16-m\\ge 0\\Leftrightarrow m\\le 16$<br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Viet, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=8\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{x}_{1}}.{{x}_{2}}=m\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{aligned} \\right.$<br\/>Theo gi\u1ea3 thi\u1ebft ${{x}_{1}}-{{x}_{2}}=2$ . K\u1ebft h\u1ee3p v\u1edbi (1), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<br\/> $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=8 \\\\ & {{x}_{1}}-{{x}_{2}}=2 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2{{x}_{1}}=10 \\\\ & {{x}_{1}}-{{x}_{2}}=2 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{1}}=5 \\\\ & {{x}_{2}}=3 \\\\ \\end{aligned} \\right.$<br\/> Thay ${{x}_{1}}=5;{{x}_{2}}=3$ v\u00e0o (2), ta c\u00f3: $5.3=m$ $\\Leftrightarrow m=15$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady $m=15$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ th\u1ecfa m\u00e3n ${{x}_{1}}-{{x}_{2}}=2$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $15.$<\/span>"}]}],"id_ques":887},{"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":[[["12"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-8x+m=0.$ T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ th\u1ecfa m\u00e3n ${{x}_{1}}=3{{x}_{2}}$ <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m=$_input_ ","hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-8x+m=0$ c\u00f3 $\\Delta '=16-m$<br\/> Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $\\Leftrightarrow \\Delta '\\ge 0\\Leftrightarrow 16-m\\ge 0\\Leftrightarrow m\\le 16$<br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Viet, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=8\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{x}_{1}}.{{x}_{2}}=m\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{aligned} \\right.$<br\/> Thay ${{x}_{1}}=3{{x}_{2}}$ v\u00e0o (1), ta c\u00f3:<br\/> $3{{x}_{2}}+{{x}_{2}}=8\\Leftrightarrow 4{{x}_{2}}=8\\Leftrightarrow {{x}_{2}}=2$ <br\/>Suy ra ${{x}_{1}}=3.2=6$ <br\/>Thay ${{x}_{1}}=6;{{x}_{2}}=2$ v\u00e0o (2), ta c\u00f3: $6.2=m$$\\Leftrightarrow m=12$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady $m =12$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ th\u1ecfa m\u00e3n ${{x}_{1}}=3{{x}_{2}}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $12.$<\/span>"}]}],"id_ques":888},{"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 $x^2+2x+m=0$ c\u00f3 hai nghi\u1ec7m $x_1;x_2$ ph\u00e2n bi\u1ec7t th\u1ecfa m\u00e3n $x_{1}^{2}+x_{2}^{2}=1$ ","select":["A. $m=\\dfrac{3}{2}$","B. $m=-\\dfrac{3}{2}$ ","C. $m=\\dfrac{1}{2}$ ","D. Kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $m$"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $\\Delta '=1-m$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $\\Leftrightarrow \\Delta '>0 \\Leftrightarrow 1-m>0\\Leftrightarrow m<1$ (*)<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=-2 \\\\ & {{x}_{1}}{{x}_{2}}=m \\\\ \\end{aligned} \\right.$ <br\/>Theo gi\u1ea3 thi\u1ebft: $x_{1}^{2}+x_{2}^{2}=1$<br\/>$\\begin{aligned} & \\Leftrightarrow {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}}=1 \\\\ & \\Leftrightarrow 4-2m=1 \\\\ & \\Leftrightarrow 2m=3 \\\\ \\end{aligned}$<br\/>$\\Leftrightarrow m=\\dfrac{3}{2}$ (kh\u00f4ng th\u1ecfa m\u00e3n (*))<br\/>V\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $m$ th\u1ecfa m\u00e3n y\u00eau c\u1ea7u c\u1ee7a \u0111\u1ec1 b\u00e0i.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":4}]}],"id_ques":889},{"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":" Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-mx+2m=0$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $-1.$ Nghi\u1ec7m c\u00f2n l\u1ea1i l\u00e0 ","select":["A. $-1$","B. $6$ ","C. $\\dfrac{2}{3}$ ","D. $-\\dfrac{2}{3}$"],"hint":"T\u00ecm $m$ tr\u01b0\u1edbc, sau \u0111\u00f3 \u00e1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t \u0111\u1ec3 t\u00ecm nghi\u1ec7m c\u00f2n l\u1ea1i.","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-mx+2m=0$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $-1$ n\u00ean $a-b+c=0$$\\Leftrightarrow 1+m+2m=0\\Leftrightarrow m=-\\dfrac{1}{3}$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi \u2013\u00e9t, ta c\u00f3: ${{x}_{1}}.{{x}_{2}}=2m$ <br\/>Suy ra <br\/>$\\begin{align} & \\left( -1 \\right){{x}_{2}}=2.\\left( -\\dfrac{1}{3} \\right)\\Leftrightarrow {{x}_{2}}=\\dfrac{2}{3} \\\\ & \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":4}]}],"id_ques":890}],"lesson":{"save":0,"level":1}}