{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["x-1"],["x+1","1+x"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ C=1$$+\\left( \\dfrac{x+1}{{{x}^{3}}+1}-\\dfrac{1}{x-{{x}^{2}}-1}-\\dfrac{2}{x+1} \\right):\\dfrac{{{x}^{3}}-2{{x}^{2}}}{{{x}^{3}}-{{x}^{2}}+x} $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $C$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $C = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{-1;0;2\\}$ <br\/> Ta c\u00f3: <br\/> $ C=1$$+\\left( \\dfrac{x+1}{{{x}^{3}}+1}-\\dfrac{1}{x-{{x}^{2}}-1}-\\dfrac{2}{x+1} \\right):\\dfrac{{{x}^{3}}-2{{x}^{2}}}{{{x}^{3}}-{{x}^{2}}+x} $<br\/>$ =1$$+\\left[ \\dfrac{x+1}{{{x}^{3}}+1}+\\dfrac{x+1}{{{x}^{3}}+1}-\\dfrac{2\\left( {{x}^{2}}-x+1 \\right)}{{{x}^{3}}+1} \\right]:\\dfrac{{{x}^{2}}\\left( x-2 \\right)}{x\\left( {{x}^{2}}-x+1 \\right)} $<br\/>$ =1$$+\\dfrac{x+1+x+1-2\\left( {{x}^{2}}-x+1 \\right)}{{{x}^{3}}+1}:\\dfrac{{{x}^{2}}\\left( x-2 \\right)}{x\\left( {{x}^{2}}-x+1 \\right)} $<br\/>$ =1$$+\\dfrac{-2{{x}^{2}}+4x}{\\left( x+1 \\right)\\left( {{x}^{2}}-x+1 \\right)}\\cdot \\dfrac{x\\left( {{x}^{2}}-x+1 \\right)}{{{x}^{2}}\\left( x-2 \\right)} $<br\/>$ =1$$+\\dfrac{-2x\\left( x-2 \\right)x\\left( {{x}^{2}}-x+1 \\right)}{\\left( x+1 \\right)\\left( {{x}^{2}}-x+1 \\right){{x}^{2}}\\left( x-2 \\right)} $<br\/>$ =1+\\dfrac{-2}{x+1} $<br\/>$ =\\dfrac{x+1-2}{x+1} $<br\/>$ =\\dfrac{x-1}{x+1} $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 c\u00f3 k\u1ebft qu\u1ea3 l\u00e0 $\\dfrac{x-1}{x+1}$. <\/span><\/span> "}]}],"id_ques":241},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank_random","correct":[[["-3"],["-2"],["1"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ C=1$$+\\left( \\dfrac{x+1}{{{x}^{3}}+1}-\\dfrac{1}{x-{{x}^{2}}-1}-\\dfrac{2}{x+1} \\right):\\dfrac{{{x}^{3}}-2{{x}^{2}}}{{{x}^{3}}-{{x}^{2}}+x} $<br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> T\u00ecm $x\\in \\mathbb{Z}$ \u0111\u1ec3 $C\\in \\mathbb{Z}$ <br\/> \u0110\u00e1p \u00e1n: $x\\in \\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, ta c\u00f3: $C=\\dfrac{x-1}{x+1}$<br\/> $\\Rightarrow C =\\dfrac{x+1-2}{x+1}$$=1-\\dfrac{2}{x+1}$ <br\/> \u0110\u1ec3 $C\\in \\mathbb{Z}$ th\u00ec $\\dfrac{2}{x+1}\\in \\mathbb{Z}$ <br\/> Khi \u0111\u00f3 $x+1\\in \u01af\\left( 2 \\right)$ , m\u00e0 $\u01af\\left( 2 \\right)=\\left\\{ -2; -1;1;2 \\right\\}$ <br\/> Ta c\u00f3 b\u1ea3ng sau:<br\/> <table> <tr> <th>$x+1$<\/th> <th>$-2$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <th>$2$<\/th> <\/tr> <tr> <td>$x$<\/td> <td>$-3$<\/td> <td>$-2$<\/td> <td>$0$ (lo\u1ea1i)<\/td> <td>$1$<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-3; -2; 1.$ <\/span><\/span> "}]}],"id_ques":242},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["-3"]]],"list":[{"point":10,"width":40,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ C=1$$+\\left( \\dfrac{x+1}{{{x}^{3}}+1}-\\dfrac{1}{x-{{x}^{2}}-1}-\\dfrac{2}{x+1} \\right):\\dfrac{{{x}^{3}}-2{{x}^{2}}}{{{x}^{3}}-{{x}^{2}}+x} $<br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> V\u1edbi $\\left|x-\\dfrac{3}{4}\\right|=\\dfrac{5}{4}$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $C$ l\u00e0 _input_ <\/span> ","hint":" T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x$ t\u1eeb $\\left|x-\\dfrac{3}{4}\\right|=\\dfrac{5}{4}$<br\/> Thay gi\u00e1 tr\u1ecb c\u1ee7a $x$ t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o bi\u1ec3u th\u1ee9c $C$ \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, ta c\u00f3: $C=\\dfrac{x-1}{x+1}$ <br\/> $\\left|x-\\dfrac{3}{4}\\right|=\\dfrac{5}{4}$ <br\/> $\\Rightarrow\\left[ \\begin{aligned} & x-\\dfrac{3}{4}=\\dfrac{5}{4} \\\\ & x-\\dfrac{3}{4}=-\\dfrac{5}{4} \\\\ \\end{aligned} \\right.\\Rightarrow \\left[ \\begin{aligned} & x=2\\,\\,\\,\\left( \\text{lo\u1ea1i} \\right) \\\\ & x=-\\dfrac{1}{2}\\,\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ \\end{aligned} \\right.$ <br\/> Thay $x=\\dfrac{-1}{2}$ v\u00e0o $C$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c: <br\/> $C=\\dfrac{-\\dfrac{1}{2}-1}{-\\dfrac{1}{2}+1}=\\dfrac{\\dfrac{-3}{2}}{\\dfrac{1}{2}}=-3$ <\/span> "}]}],"id_ques":243},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ C=1$$+\\left( \\dfrac{x+1}{{{x}^{3}}+1}-\\dfrac{1}{x-{{x}^{2}}-1}-\\dfrac{2}{x+1} \\right):\\dfrac{{{x}^{3}}-2{{x}^{2}}}{{{x}^{3}}-{{x}^{2}}+x} $<br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> \u0110\u1ec3 $C = \\dfrac{1}{2}$ th\u00ec $x =$ _input_ <\/span> ","explain":"<span class='basic_left'><br\/> Theo c\u00e2u 1 tr\u00ean, ta c\u00f3: $C=\\dfrac{x-1}{x+1}$ <br\/> $C=\\dfrac{1}{2} \\Leftrightarrow$ $ \\dfrac{x-1}{x+1}=\\dfrac{1}{2} $ <br\/> $ \\Leftrightarrow 2(x-1)=x+1 $ <br\/> $ \\Leftrightarrow 2x-2=x+1 $ <br\/> $ \\Leftrightarrow x=3 $ (th\u1ecfa m\u00e3n)<\/span> "}]}],"id_ques":244},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"],["-2"]]],"list":[{"point":10,"width":40,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/9.jpg' \/><\/center> T\u00ecm c\u00e1c h\u1ec7 s\u1ed1 $a, b$ sao cho $\\dfrac{5x-2}{{{x}^{2}}+x-20}=\\dfrac{a}{x+5}-\\dfrac{b}{x-4}$ v\u1edbi $x\\ne \\{-5;4\\}$ <br\/> \u0110\u00e1p \u00e1n: $a =$ _input_ ; $b =$ _input_ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c v\u00e0 r\u00fat g\u1ecdn $\\dfrac{a}{x+5}-\\dfrac{b}{x-4}$ \u0111\u01b0a v\u1ec1 c\u00f9ng m\u1eabu v\u1edbi v\u1ebf tr\u00e1i.<br\/> <b> B\u01b0\u1edbc 2:<\/b> \u0110\u1ed3ng nh\u1ea5t t\u1eed th\u1ee9c c\u1ee7a v\u1ebf tr\u00e1i v\u00e0 v\u1ebf ph\u1ea3i, \u0111\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 \u0111\u1ec3 t\u00ecm $a, b.$ <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne \\{-5; 4\\}$ <br\/> Ta c\u00f3:<br\/> $\\begin{align} & \\text{V\u1ebf ph\u1ea3i}=\\dfrac{a}{x+5}-\\dfrac{b}{x-4} \\\\ & =\\dfrac{a\\left( x-4 \\right)-b\\left( x+5 \\right)}{\\left( x+5 \\right)\\left( x-4 \\right)} \\\\ & =\\dfrac{a\\,x-4a-bx-5b}{\\left( x+5 \\right)\\left( x-4 \\right)} \\\\ & =\\dfrac{\\left( a-b \\right)x-4a-5b}{\\left( x+5 \\right)\\left( x-4 \\right)} \\\\ & \\text{V\u1ebf tr\u00e1i}=\\dfrac{5x-2}{{{x}^{2}}+x-20} \\\\ & =\\dfrac{5x-2}{{{x}^{2}}+5x-4x-20} \\\\ & =\\dfrac{5x-2}{x\\left( x+5 \\right)-4\\left( x+5 \\right)} \\\\ & =\\dfrac{5x-2}{\\left( x+5 \\right)\\left( x-4 \\right)} \\\\ \\end{align}$ <br\/> \u0110\u1ed3ng nh\u1ea5t hai t\u1eed th\u1ee9c: $5x-2\\equiv \\left( a-b \\right)x-4a-5b$, ta \u0111\u01b0\u1ee3c:<br\/> $\\left\\{ \\begin{aligned} & a-b=5 \\\\ & -4a-5b=-2 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & a=3 \\\\ & b=-2 \\\\ \\end{aligned} \\right.$<\/span> "}]}],"id_ques":245},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-1"],["-5"],["-3"],["1"]]],"list":[{"point":10,"width":30,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/8.jpg' \/><\/center> <span class='basic_left'>T\u00ecm $x$ nguy\u00ean \u0111\u1ec3 $Q=\\dfrac{2{{x}^{2}}+3x+1}{x+2} \\in \\mathbb{Z}$ <br\/> <br\/> <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x\\in \\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$<\/span> ","hint":" Ph\u00e2n t\u00edch t\u1eed th\u1ee9c l\u00e0m xu\u1ea5t hi\u1ec7n $x+2$ \u0111\u1ec3 r\u00fat g\u1ecdn v\u1edbi m\u1eabu th\u1ee9c. ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne -2$ <br\/> Ta c\u00f3:<br\/> $\\begin{align} & Q=\\dfrac{2{{x}^{2}}+3x+1}{x+2} \\\\ & =\\dfrac{2{{x}^{2}}+4x-x+1}{x+2} \\\\ & =\\dfrac{2x\\left( x+2 \\right)-\\left( x+2 \\right)+3}{x+2} \\\\ & =\\dfrac{2x\\left( x+2 \\right)}{x+2}-\\dfrac{x+2}{x+2}+\\dfrac{3}{x+2} \\\\ & =2x-1+\\dfrac{3}{x+2} \\\\ \\end{align}$ <br\/> Do $x\\in \\mathbb{Z}\\Rightarrow 2x-1\\,\\,\\in \\mathbb{Z}$ <br\/> \u0110\u1ec3 $Q\\in \\mathbb{Z}$ th\u00ec $\\dfrac{3}{x+2}\\in \\mathbb{Z}$ <br\/> Khi \u0111\u00f3 $x+1\\in \u01af\\left( 3 \\right)$ , m\u00e0 $\u01af\\left( 3 \\right)=\\left\\{ -3;-1;1;3 \\right\\}$ <br\/> Ta c\u00f3 b\u1ea3ng sau:<br\/> <table> <tr> <th>$x+2$<\/th> <th>$-3$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <th>$3$<\/th> <\/tr> <tr> <td>$x$<\/td> <td>$-5$<\/td> <td>$-3$<\/td> <td>$-1$<\/td> <td>$1$<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-5; -3; -1; 1.$ <\/span><br\/><br\/> <b> L\u01b0u \u00fd:<\/b> Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i \u0111\u1ed1i v\u1edbi d\u1ea1ng b\u00e0i n\u00e0y l\u00e0: <br\/> \u0110\u01b0a bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng $z+\\dfrac{a}{A}$, trong \u0111\u00f3 $z\\in\\mathbb{Z}$<br\/> Ph\u00e2n th\u1ee9c $\\dfrac{a}{A}\\in\\mathbb{Z}$ th\u00ec $A$ thu\u1ed9c t\u1eadp \u01b0\u1edbc nguy\u00ean c\u1ee7a $a$ v\u1edbi $a\\in \\mathbb{Z}$ <\/span> "}]}],"id_ques":246},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2017"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/5.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{x+\\dfrac{1}{{{x}^{2}}}}{1-\\dfrac{1}{x}+\\dfrac{1}{{{x}^{2}}}}$ t\u1ea1i $x = 2016$ l\u00e0 _input_ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x = 2016$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne 0$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{x+\\dfrac{1}{{{x}^{2}}}}{1-\\dfrac{1}{x}+\\dfrac{1}{{{x}^{2}}}} \\\\ & =\\dfrac{\\dfrac{{{x}^{3}}+1}{{{x}^{2}}}}{\\dfrac{{{x}^{2}}-x+1}{{{x}^{2}}}} \\\\ & =\\dfrac{{{x}^{3}}+1}{{{x}^{2}}}\\cdot \\dfrac{{{x}^{2}}}{{{x}^{2}}-x+1} \\\\ & =\\dfrac{\\left( x+1 \\right)\\left( {{x}^{2}}-x+1 \\right){{x}^{2}}}{{{x}^{2}}\\left( {{x}^{2}}-x+1 \\right)} \\\\ & =x+1 \\\\ \\end{align}$ <br\/> Thay $x = 2016$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $2016+1=2017$ <\/span> "}]}],"id_ques":247},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/4.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{{{\\left( 2x+3 \\right)}^{2}}+2\\left( 4{{x}^{2}}-9 \\right)+{{\\left( 2x-3 \\right)}^{2}}}{{{\\left( 2x-3 \\right)}^{2}}-2\\left( 4{{x}^{2}}-9 \\right)+{{\\left( 2x+3 \\right)}^{2}}}$ l\u00e0: ","select":["A. $\\dfrac{4x}{9}$ ","B. $\\dfrac{4x^2}{9}$","C. $\\dfrac{x^2}{9}$","D. $\\dfrac{4}{9}$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u01b0a t\u1eed th\u1ee9c v\u00e0 m\u1eabu th\u1ee9c th\u00e0nh b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t hi\u1ec7u ho\u1eb7c m\u1ed9t t\u1ed5ng. <br\/><br\/><span class='basic_green'> Gi\u1ea3i<\/span><br\/><span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{{{\\left( 2x+3 \\right)}^{2}}+2\\left( 4{{x}^{2}}-9 \\right)+{{\\left( 2x-3 \\right)}^{2}}}{{{\\left( 2x-3 \\right)}^{2}}-2\\left( 4{{x}^{2}}-9 \\right)+{{\\left( 2x+3 \\right)}^{2}}} \\\\ & =\\dfrac{{{\\left( 2x+3 \\right)}^{2}}+2\\left( 2x+3 \\right)\\left( 2x-3 \\right)+{{\\left( 2x-3 \\right)}^{2}}}{{{\\left( 2x-3 \\right)}^{2}}-2\\left( 2x+3 \\right)\\left( 2x-3 \\right)+{{\\left( 2x+3 \\right)}^{2}}} \\\\ & =\\dfrac{{{\\left[ \\left( 2x+3 \\right)+\\left( 2x-3 \\right) \\right]}^{2}}}{{{\\left[ \\left( 2x-3 \\right)-\\left( 2x+3 \\right) \\right]}^{2}}} \\\\ & =\\dfrac{{{\\left( 4x \\right)}^{2}}}{{{\\left( -6 \\right)}^{2}}} \\\\ & =\\dfrac{16{{x}^{2}}}{36} \\\\ & =\\dfrac{4x^2}{9} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span> ","column":2}]}],"id_ques":248},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["4x-3"],["x+1","1+x"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/3.jpg' \/><\/center> <span class='basic_left'> T\u00ecm $P(x)$ \u0111\u1ec3 $ \\dfrac{4{{x}^{2}}-7x+3}{{{x}^{2}}-1}$$=\\dfrac{P\\left( x \\right)}{{{x}^{2}}+2x+1} $ <br\/> <br\/> \u0110\u00e1p \u00e1n: $P(x)=(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\pm 1$ <br\/> Ta c\u00f3:<br\/> $ \\dfrac{4{{x}^{2}}-7x+3}{{{x}^{2}}-1}$$=\\dfrac{P\\left( x \\right)}{{{x}^{2}}+2x+1} $<br\/>$\\Leftrightarrow \\dfrac{4{{x}^{2}}-4x-3x+3}{{{x}^{2}}-1}$$=\\dfrac{P\\left( x \\right)}{{{\\left( x+1 \\right)}^{2}}} $<br\/>$\\Leftrightarrow \\dfrac{4x\\left( x-1 \\right)-3\\left( x-1 \\right)}{{{x}^{2}}-1}$$=\\dfrac{P\\left( x \\right)}{{{\\left( x+1 \\right)}^{2}}} $<br\/>$\\Leftrightarrow \\dfrac{\\left( x-1 \\right)\\left( 4x-3 \\right)}{\\left( x+1 \\right)\\left( x-1 \\right)}$$=\\dfrac{P\\left( x \\right)}{{{\\left( x+1 \\right)}^{2}}} $<br\/>$\\Leftrightarrow \\dfrac{4x-3}{x+1}$$=\\dfrac{P\\left( x \\right)}{{{\\left( x+1 \\right)}^{2}}} $<br\/>$\\Leftrightarrow \\left( x+1 \\right)P\\left( x \\right)$$=\\left( 4x-3 \\right){{\\left( x+1 \\right)}^{2}} $<br\/>$ \\Leftrightarrow P\\left( x \\right)$$=\\dfrac{\\left( 4x-3 \\right){{\\left( x+1 \\right)}^{2}}}{x+1} $<br\/>$\\Leftrightarrow P\\left( x \\right)=\\left( 4x-3 \\right)\\left( x+1 \\right) $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 c\u00f3 $P=\\left( 4x-3 \\right)\\left( x+1 \\right)$. <\/span><\/span> "}]}],"id_ques":249},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["x"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv3/img\/2.jpg' \/><\/center> <span class='basic_left'> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/> $\\dfrac{x-\\dfrac{1}{{{x}^{2}}}}{1+\\dfrac{1}{x}+\\dfrac{1}{{{x}^{2}}}}+1$$=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne 0$<br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{x-\\dfrac{1}{{{x}^{2}}}}{1+\\dfrac{1}{x}+\\dfrac{1}{{{x}^{2}}}} +1 \\\\ & =\\dfrac{\\dfrac{{{x}^{3}}-1}{{{x}^{2}}}}{\\dfrac{{{x}^{2}}+x+1}{{{x}^{2}}}} +1 \\\\ & =\\dfrac{{{x}^{3}}-1}{x^2}\\cdot \\dfrac{x^2}{{{x}^{2}}+x+1} + 1 \\\\ & =\\dfrac{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)x^2}{x^2\\left( {{x}^{2}}-x+1 \\right)} + 1 \\\\ & =x- 1 + 1 \\\\ & = x\\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $x$. <\/span><\/span> "}]}],"id_ques":250}],"lesson":{"save":0,"level":3}}