{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["-9"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":" T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $A=\\sqrt{\\dfrac{{{x}^{2}}}{9}+2x+10}$ <br\/>\u0110\u00e1p s\u1ed1: $A_{\\min}=$_input_ khi $x =$ _input_","hint":"S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c d\u01b0\u1edbi d\u1ea5u c\u0103n v\u1ec1 d\u1ea1ng ${{\\left[ f\\left( x \\right) \\right]}^{2}}+a$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $A$ v\u1ec1 d\u1ea1ng: $A=\\sqrt{[f(x)]^2 + a}$ <br\/><b>B\u01b0\u1edbc 2:<\/b> \u0110\u00e1nh gi\u00e1 v\u00e0 t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $A$.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $A=\\sqrt{\\dfrac{{{x}^{2}}}{9}+2x+10}$$\\,=\\sqrt{{{\\left( \\dfrac{x}{3} \\right)}^{2}}+2.\\dfrac{x}{3}.3+{{3}^{2}}+1}$$\\,=\\sqrt{{{\\left( \\dfrac{x}{3}+3 \\right)}^{2}}+1}$ <br\/>V\u00ec ${{\\left( \\dfrac{x}{3}+3 \\right)}^{2}}\\ge \\,0\\,\\forall x\\,\\Rightarrow \\sqrt{{{\\left( \\dfrac{x}{3}+3 \\right)}^{2}}+1}\\ge \\sqrt{1}=1\\,\\forall x$ <br\/>Suy ra, ${{A}_{\\min }}=1\\,\\Leftrightarrow {{\\left( \\dfrac{x}{3}+3 \\right)}^{2}}=0\\,\\Leftrightarrow x=-9$ <\/span> <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 $A_{\\min}=1$ khi $x = -9$<\/span> <\/span><\/span> "}]}],"id_ques":501},{"time":24,"part":[{"title":"","title_trans":"","temp":"fill_the_blank","correct":[[["3"],["-2"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $A=\\sqrt{-a^2-4a+5}$<br\/>\u0110\u00e1p \u00e1n: $A_{\\max}=$_input_ khi $a =$ _input_","hint":"S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c d\u01b0\u1edbi d\u1ea5u c\u0103n v\u1ec1 d\u1ea1ng $b-{{\\left[ f\\left( a \\right) \\right]}^{2}}$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $A$ v\u1ec1 d\u1ea1ng: $A=\\sqrt{b - [f(a)]^2}$ <br\/><b>B\u01b0\u1edbc 2:<\/b> \u0110\u00e1nh gi\u00e1 v\u00e0 t\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a $A$.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/>$A=\\sqrt{-{{a}^{2}}-4a+5}\\,$$=\\sqrt{-\\left( {{a}^{2}}+4a+4 \\right)+9}$$\\,=\\sqrt{9-{{\\left( a + 2 \\right)}^{2}}}$ <br\/>Nh\u1eadn x\u00e9t: ${{\\left( a+2 \\right)}^{2}}\\ge 0\\,\\,\\,\\,\\forall a$<br\/>$\\Rightarrow 9-{{\\left( a+2 \\right)}^{2}}\\le 9\\,\\,\\forall a$<br\/>$\\Rightarrow A\\le\\sqrt 9\\,\\forall a$ <br\/>V\u1eady ${{A}_{\\text{max}}}=\\sqrt{9}=3\\Leftrightarrow {{\\left( a+2 \\right)}^{2}}=0$$\\,\\Leftrightarrow a=-2$ <br\/> <span class='basic_pink'> V\u1eady $A_{\\max}= 3$ khi $a = -2$<\/span> <\/span> "}]}],"id_ques":502},{"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"],["3"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c: $A=\\sqrt{7+4\\sqrt{3}}-\\sqrt{13-4\\sqrt{3}}$<br\/>\u0110\u00e1p s\u1ed1: $A = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}-\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c d\u01b0\u1edbi d\u1ea5u c\u0103n v\u1ec1 b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t t\u1ed5ng ho\u1eb7c b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t hi\u1ec7u","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(A \\pm B)^2$<br\/> <b>B\u01b0\u1edbc 2:<\/b> Khai c\u0103n, t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $A.$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\begin{aligned} A &= \\sqrt{7+4\\sqrt{3}}-\\sqrt{13-4\\sqrt{3}} \\\\ &=\\sqrt{2^2+4\\sqrt{3} + (\\sqrt{3})^2}-\\sqrt{(2\\sqrt{3})^2 - 4\\sqrt{3} + 1} \\\\ & =\\sqrt{{{\\left( 2+\\sqrt{3} \\right)}^{2}}}-\\sqrt{{{\\left( 2\\sqrt{3}-1 \\right)}^{2}}} \\\\ & =\\left| 2+\\sqrt{3} \\right|-\\left| 2\\sqrt{3}-1 \\right| \\\\ & =2+\\sqrt{3}- 2\\sqrt{3}+1\\,\\,(\\text{V\u00ec}\\,\\,2\\sqrt{3}>1)\\\\ & =3-\\sqrt{3} \\\\ \\end{aligned}$<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 $A=3-\\sqrt{3}$ <\/span><br\/><b><i>L\u01b0u \u00fd:<\/i><\/b><br\/> V\u1edbi m\u1ecdi s\u1ed1 th\u1ef1c $a$, ta c\u00f3 $\\sqrt{a^2}=|a|=\\left\\{ \\begin{align} & a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\, a\\ge {0} \\\\ & -a\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,a< {0} \\\\\\end{align} \\right.$. <\/span> "}]}],"id_ques":503},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"$\\sqrt{a+2\\sqrt{a-1}}+\\sqrt{a-2\\sqrt{a-1}}=2$ v\u1edbi $1 < a < 2$ ","select":["A. \u0110\u00fang ","B. Sai "],"hint":" Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(A \\pm B)^2$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(A \\pm B)^2$<br\/> <b>B\u01b0\u1edbc 2:<\/b> Khai c\u0103n, t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $A$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'> Ta c\u00f3:<br\/>$ \\sqrt{a+2\\sqrt{a-1}}+\\sqrt{a-2\\sqrt{a-1}} $<br\/> <br\/>$=\\sqrt{a-1+2\\sqrt{a-1}+1}+$$\\sqrt{a-1-2\\sqrt{a-1}+1}$ <br\/> <br\/> $=\\sqrt{{{\\left( \\sqrt{a-1}+1 \\right)}^{2}}}+$$\\sqrt{{{\\left( \\sqrt{a-1}-1 \\right)}^{2}}}$<br\/> <br\/>$=\\left| \\sqrt{a-1}+1 \\right|+\\left| \\sqrt{a-1}-1 \\right| $<br\/> <br\/> $ =\\sqrt{a-1}+1-\\sqrt{a-1}+1$$ \\,\\,\\,(\\text{V\u00ec}\\,\\,1 < a < 2 \\,\\,\\text{n\u00ean}\\sqrt{a-1}-1<0) $<br\/> <br\/>$ =2$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A <\/span><br\/><b><i>L\u01b0u \u00fd:<\/i><\/b><br\/> V\u1edbi m\u1ecdi s\u1ed1 th\u1ef1c $a$, ta c\u00f3 $\\sqrt{a^2}=|a|=\\left\\{ \\begin{align} & a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\, a\\ge {0} \\\\ & -a\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,a< {0} \\\\\\end{align} \\right.$. <\/span>","column":2}]}],"id_ques":504},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"V\u1edbi $x\\ge 3$, $A=\\sqrt{{{x}^{2}}-6x+9}-2x+1$ c\u00f3 k\u1ebft qu\u1ea3 r\u00fat g\u1ecdn l\u00e0: ","select":["A. $-3x + 2$ ","B. $2 - x$ ","C. $-x -2$","D. $x - 2$"],"hint":"","explain":"<span class='basic_left'><span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Khai c\u0103n<br\/><b>B\u01b0\u1edbc 2:<\/b> Thu g\u1ecdn $A$<\/span> <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{aligned}A &= \\sqrt{{{x}^{2}}-6x+9}-2x+1 \\\\ & =\\sqrt{{{\\left( x-3 \\right)}^{2}}}-2x+1 \\\\ & =\\left| x-3 \\right|-2x+1 \\\\ & =x-3-2x+1 \\,\\,\\,(\\text{V\u00ec}\\, x > 3\\, \\text{n\u00ean}\\,\\,|x-3|=x-3 )\\\\ & =-x-2 \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span> <br\/><\/span>","column":2}]}],"id_ques":505},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-1"]]],"list":[{"point":10,"width":60,"type_input":"","ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $D=\\dfrac{\\sqrt{{{x}^{2}}+10x+25}}{x+5} \\,\\,\\, \\text{v\u1edbi}\\,\\, x < -5$<br\/>\u0110\u00e1p \u00e1n:$D = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(A \\pm B)^2$<br\/><b>B\u01b0\u1edbc 2:<\/b> Khai c\u0103n, thu g\u1ecdn $A$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>V\u00ec $x<-5\\Leftrightarrow \\left( x+5 \\right)<0\\,$$\\Rightarrow \\left| x+5 \\right|=-\\left( x+5 \\right)$ <br\/>Ta c\u00f3:<br\/> $D=\\dfrac{\\sqrt{{{x}^{2}}+10x+25}}{x+5}\\,$<br\/>$=\\dfrac{\\sqrt{{{\\left( x+5 \\right)}^{2}}}}{x+5}\\,$<br\/>$=\\dfrac{\\left| x+5 \\right|}{x+5}\\,$<br\/>$=\\dfrac{-\\left( x+5 \\right)}{x+5}$<br\/>$=-1$<br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $ -1$ <\/span><\/span> "}]}],"id_ques":506},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4","49"],["49","4"]]],"list":[{"point":10,"width":100,"type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x-9\\sqrt{x}+14=0$ <br\/>\u0110\u00e1p s\u1ed1:$\\left[ \\begin{array}{} x= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ x= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\end{array} \\right.$","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch \u0111\u1ec3 ph\u00e2n t\u00edch \u0111a th\u1ee9c v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u00f3 ngh\u0129a c\u1ee7a c\u0103n th\u1ee9c<br\/> <b>B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed <br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ecm $x$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ge 0$<br\/>$\\begin{aligned} &x-9\\sqrt{x}+14=0 \\\\ \\Leftrightarrow &x-2\\sqrt{x}-7\\sqrt{x}+14=0 \\\\ \\Leftrightarrow & \\sqrt{x} (\\sqrt{x} - 2) - 7(\\sqrt{x} - 2) = 0 \\\\ \\Leftrightarrow &\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}-7 \\right)=0 \\\\ \\end{aligned}$ <br\/>$\\Leftrightarrow \\left[ \\begin{aligned} & \\sqrt{x}=2 \\\\ & \\sqrt{x}=7 \\\\ \\end{aligned} \\right.$<br\/>$\\Leftrightarrow \\left[ \\begin{aligned} & x=4 \\\\ & x=49 \\\\ \\end{aligned} \\right.$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\left\\{ 4;49\\right\\}$ <br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4$ ho\u1eb7c $49$ <\/span><\/span>"}]}],"id_ques":507},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-2"],["2"]]],"list":[{"point":10,"width":100,"type_input":"","ques":"Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $\\sqrt{x+2}>x$ <br\/>\u0110\u00e1p \u00e1n: _input_$\\,<\\, x\\,<$_input_","hint":"X\u00e9t hai tr\u01b0\u1eddng h\u1ee3p $x<0$ v\u00e0 $x\\ge0$ \u0111\u1ec3 gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec1 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ngh\u0129a<br\/><b>B\u01b0\u1edbc 2:<\/b> X\u00e9t tr\u01b0\u1eddng h\u1ee3p $x<0$<br\/> <b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi $x\\ge 0$<br\/><b>B\u01b0\u1edbc 4:<\/b> K\u1ebft lu\u1eadn t\u1eadp nghi\u1ec7m<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ge-2$<br\/>Tr\u01b0\u1eddng h\u1ee3p 1: V\u1edbi $x<0$<br\/>Ta c\u00f3:$\\sqrt{x+2}>x$<br\/>V\u00ec $\\sqrt{x+2}\\ge 0\\,\\,\\,\\forall x\\ge -2$ v\u00e0 $x<0$ n\u00ean $\\sqrt{x+2}>x \\,\\,\\,\\,$ v\u1edbi $-2 \\le x<0$<br\/>V\u1eady $-2\\le x<0.$<br\/>Tr\u01b0\u1eddng h\u1ee3p 2: V\u1edbi $x\\ge 0$<br\/>Ta c\u00f3:$\\sqrt{x+2}>x\\Leftrightarrow x+2>{{x}^{2}}$<br\/>$\\,\\Leftrightarrow {{x}^{2}}-x-2<0$<br\/>$\\,\\Leftrightarrow \\left( x-2 \\right)\\left( x+1 \\right)<0$ <br\/>+)$\\left\\{ \\begin{aligned} & x-2>0 \\\\ & x+1<0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x>2 \\\\ & x<-1 \\\\ \\end{aligned} \\right.$ (lo\u1ea1i) <br\/>+)$\\left\\{ \\begin{aligned} & x-2<0 \\\\ & x+1>0 \\\\ \\end{aligned} \\right.$$\\Leftrightarrow \\left\\{ \\begin{aligned} & x<2 \\\\ & x>-1 \\\\ \\end{aligned} \\right.$$\\,\\Leftrightarrow -1 < x < 2$ <br\/>V\u1eady nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $-2 < x < 2$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-2$ v\u00e0 $2$ <\/span><br\/><i>Nh\u1eadn x\u00e9t:<\/i> V\u1edbi b\u00e0i to\u00e1n gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 d\u1ea1ng: $\\sqrt{f(x)}>g(x)$, ta th\u1ef1c hi\u1ec7n:<br\/><b>B\u01b0\u1edbc 1:<\/b>T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh.<br\/><b>B\u01b0\u1edbc 2:<\/b> Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi tr\u01b0\u1eddng h\u1ee3p $g(x)<0$.<br\/><b>B\u01b0\u1edbc 3:<\/b>Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi tr\u01b0\u1eddng h\u1ee3p $g(x)\\ge0$.<br\/><b>B\u01b0\u1edbc 4:<\/b> K\u1ebft lu\u1eadn t\u1eadp nghi\u1ec7m <\/span>"}]}],"id_ques":508},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"Nh\u1eefng gi\u00e1 tr\u1ecb nguy\u00ean n\u00e0o c\u1ee7a $x$ \u0111\u1ec3 bi\u1ec3u th\u1ee9c $\\sqrt{\\dfrac{2x-1}{2-x}}$ c\u00f3 ngh\u0129a? <br\/> \u0110\u00e1p s\u1ed1: $x=$_input_","hint":"$\\sqrt{A}$ c\u00f3 ngh\u0129a $\\Leftrightarrow A \\ge 0$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $\\sqrt A$ c\u00f3 ngh\u0129a.<br\/><b>B\u01b0\u1edbc 2:<\/b> Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh:$\\dfrac{2x-1}{2-x}\\ge 0$ <br\/>Tr\u01b0\u1eddng h\u1ee3p 1: <br\/>$\\left\\{ \\begin{aligned} & 2x-1\\ge 0 \\\\ & 2-x>0 \\\\\\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge \\dfrac{1}{2} \\\\ & x<2 \\\\ \\end{aligned} \\right.$$\\,\\Leftrightarrow \\dfrac{1}{2}\\le x<2$ <br\/>Tr\u01b0\u1eddng h\u1ee3p 2:<br\/>$\\left\\{ \\begin{aligned} & 2x-1 \\le 0 \\\\ & 2-x<0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x \\le \\dfrac{1}{2} \\\\ & x>2 \\\\ \\end{aligned} \\right.$ (lo\u1ea1i)<br\/>V\u1eady v\u1edbi $\\dfrac{1}{2}\\le x<2$ th\u00ec bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a<br\/>Khi \u0111\u00f3, gi\u00e1 tr\u1ecb nguy\u00ean c\u1ee7a $x$ t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 $x=1$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$<\/span><\/span>"}]}],"id_ques":509},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"T\u00ecm $x$ \u0111\u1ec3 bi\u1ec3u th\u1ee9c $\\sqrt{-{{x}^{2}}+2x-1}$ c\u00f3 ngh\u0129a. <br\/>\u0110\u00e1p s\u1ed1: $x=$_input_","hint":"$\\sqrt{A}$ c\u00f3 ngh\u0129a $\\Leftrightarrow A \\ge 0$ ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $\\sqrt{-{{x}^{2}}+2x-1}$ l\u00e0:<br\/>$-{{x}^{2}}+2x-1\\ge 0$<br\/>$\\,\\Leftrightarrow -{{\\left( x-1 \\right)}^{2}}\\ge 0$<br\/>$\\,\\Leftrightarrow {{\\left( x-1 \\right)}^{2}}\\le 0$ (1)<br\/>M\u00e0 ${{\\left( x-1 \\right)}^{2}}\\ge 0\\,\\,\\forall x$ (2)<br\/>T\u1eeb (1) v\u00e0 (2) suy ra $x-1=0\\Leftrightarrow x=1$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$<\/span><\/span>"}]}],"id_ques":510}],"lesson":{"save":0,"level":3}}