{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["1"],["y"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/5.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ P=\\dfrac{x}{xy-2{{y}^{2}}}$$-\\dfrac{2}{{{x}^{2}}+x-2xy-2y}\\cdot \\left( 1+\\dfrac{3x+{{x}^{2}}}{3+x} \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $P$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $P = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{-3; -1\\}; y\\ne 0; x\\ne 2y$ <br\/> Ta c\u00f3: <br\/> $ P=\\dfrac{x}{xy-2{{y}^{2}}}$$-\\dfrac{2}{{{x}^{2}}+x-2xy-2y}\\cdot \\left( 1+\\dfrac{3x+{{x}^{2}}}{3+x} \\right) $<br\/>$ =\\dfrac{x}{y\\left( x-2y \\right)}$$-\\dfrac{2}{x\\left( x+1 \\right)-2y\\left( x+1 \\right)}\\cdot \\left[ 1+\\dfrac{x\\left( 3+x \\right)}{3+x} \\right] $<br\/>$ =\\dfrac{x}{y\\left( x-2y \\right)}$$-\\dfrac{2}{\\left( x+1 \\right)\\left( x-2y \\right)}\\cdot \\left( 1+x \\right) $<br\/>$ =\\dfrac{x}{y\\left( x-2y \\right)}$$-\\dfrac{2\\left( 1+x \\right)}{\\left( x+1 \\right)\\left( x-2y \\right)} $<br\/>$ =\\dfrac{x}{y\\left( x-2y \\right)}-\\dfrac{2}{x-2y} $<br\/>$ =\\dfrac{x-2y}{y\\left( x-2y \\right)} $<br\/>$ =\\dfrac{1}{y} $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 c\u00f3 k\u1ebft qu\u1ea3 l\u00e0 $P=\\dfrac{1}{y}$. <\/span><\/span> "}]}],"id_ques":221},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["1"],["16"]]],"list":[{"point":5,"width":50,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/5.jpg' \/><\/center>Cho bi\u1ec3u th\u1ee9c $ P=\\dfrac{x}{xy-2{{y}^{2}}}$$-\\dfrac{2}{{{x}^{2}}+x-2xy-2y}\\cdot \\left( 1+\\dfrac{3x+{{x}^{2}}}{3+x} \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> V\u1edbi $y=16$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $P$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, ta c\u00f3 $P=\\dfrac{1}{y}$ <br\/> Thay $y=16$ v\u00e0o $P$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $P=\\dfrac{1}{y} = \\dfrac{1}{16}$<\/span> "}]}],"id_ques":222},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["-1"],["3"],["-3"],["2"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/9.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\left( {{x}^{2}}+2x-\\dfrac{11x-2}{3x+1} \\right)$$:\\left( x+1-\\dfrac{2{{x}^{2}}+x+2}{3x+1} \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $B$ l\u00e0: $x\\ne \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ v\u00e0 $x \\ne \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\pm \\sqrt{13}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ <\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $B$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & 3x+1\\ne 0 \\\\ & x+1-\\frac{2{{x}^{2}}+x+2}{3x+1}\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne \\frac{-1}{3} \\\\ & \\dfrac{{{x}^{2}}+3x-1}{3x+1}\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne \\dfrac{-1}{3} \\\\ & {{x}^{2}}+3x-1\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne \\dfrac{-1}{3} \\\\ & x\\ne \\dfrac{-3\\pm \\sqrt{13}}{2} \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 $\\dfrac{-1}{3}$ v\u00e0 $\\dfrac{-3\\pm \\sqrt{13}}{2}$ . <\/span><\/span> "}]}],"id_ques":223},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["19"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/9.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\left( {{x}^{2}}+2x-\\dfrac{11x-2}{3x+1} \\right)$$:\\left( x+1-\\dfrac{2{{x}^{2}}+x+2}{3x+1} \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> V\u1edbi $x=7$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $B$ l\u00e0: _input_<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $x\\ne \\left\\{\\dfrac{-1}{3};\\dfrac{-3\\pm \\sqrt{13}}{2}\\right\\}$ <br\/> Ta c\u00f3: <br\/> $B=\\left( {{x}^{2}}+2x-\\dfrac{11x-2}{3x+1} \\right)$$:\\left( x+1-\\dfrac{2{{x}^{2}}+x+2}{3x+1} \\right) $<br\/>$ = \\dfrac{{{x}^{2}}\\left( 3x+1 \\right)+2x\\left( 3x+1 \\right)-\\left( 11x-2 \\right)}{3x+1} $$:\\left[ \\dfrac{\\left( x+1 \\right)\\left( 3x+1 \\right)-\\left( 2{{x}^{2}}+x+2 \\right)}{3x+1} \\right] $<br\/>$ =\\dfrac{3{{x}^{3}}+{{x}^{2}}+6{{x}^{2}}+2x-11x+2}{3x+1}$$:\\dfrac{3{{x}^{2}}+x+3x+1-2{{x}^{2}}-x-2}{3x+1} $<br\/>$ =\\dfrac{3{{x}^{3}}+7{{x}^{2}}-9x+2}{3x+1}$$\\cdot \\dfrac{3x+1}{{{x}^{2}}+3x-1} $<br\/>$ =\\dfrac{3{{x}^{3}}+7{{x}^{2}}-9x+2}{{{x}^{2}}+3x-1} $ <br\/> Thay $x = 7$ v\u00e0o $B$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $ B =\\dfrac{{{3.7}^{3}}+{{7.7}^{2}}-9.7+2}{{{7}^{2}}+3.7-1}=19$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $19$. <\/span><\/span> "}]}],"id_ques":224},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank_random","correct":[[["2"],["-2"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ A=\\left( \\dfrac{1}{2-x}+\\dfrac{3x}{{{x}^{2}}-4}-\\dfrac{2}{2+x} \\right)$$:\\left( \\dfrac{{{x}^{2}}+4}{4-{{x}^{2}}}+1 \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $A$ l\u00e0: $\\left\\{ \\begin{aligned} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $A$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & 2-x\\ne 0 \\\\ & x^2-4\\ne 0 \\\\ & 2+x\\ne 0 \\\\ & \\dfrac{{{x}^{2}}+4}{4-{{x}^{2}}}+1\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 2 \\\\ & (x+2)(x-2) \\ne 0 \\\\ & x \\ne -2\\\\ & \\dfrac{8}{4-{{x}^{2}}}\\ne 0 \\\\ \\end{aligned} \\right.$ $\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 2 \\\\ & x \\ne -2 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $- 2$ v\u00e0 $2.$ <\/span><\/span> "}]}],"id_ques":225},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["-1"],["4"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ A=\\left( \\dfrac{1}{2-x}+\\dfrac{3x}{{{x}^{2}}-4}-\\dfrac{2}{2+x} \\right)$$:\\left( \\dfrac{{{x}^{2}}+4}{4-{{x}^{2}}}+1 \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $A = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $x\\ne \\{-2;2\\}$ <br\/> Ta c\u00f3: <br\/> $ A=\\left( \\dfrac{1}{2-x}+\\dfrac{3x}{{{x}^{2}}-4}-\\dfrac{2}{2+x} \\right)$$:\\left( \\dfrac{{{x}^{2}}+4}{4-{{x}^{2}}}+1 \\right) $<br\/>$=\\left[ \\dfrac{-1}{x-2}+\\dfrac{3x}{\\left( x+2 \\right)\\left( x-2 \\right)}-\\dfrac{2}{x+2} \\right]$$:\\dfrac{{{x}^{2}}+4+4-{{x}^{2}}}{4-{{x}^{2}}} $<br\/>$ =\\dfrac{-\\left( x+2 \\right)+3x-2\\left( x-2 \\right)}{\\left( x+2 \\right)\\left( x-2 \\right)}$$:\\dfrac{8}{4-{{x}^{2}}} $<br\/>$ =\\dfrac{-x-2+3x-2x+4}{{{x}^{2}}-4}$$\\cdot \\dfrac{4-{{x}^{2}}}{8} $<br\/>$ =\\dfrac{2\\left( 4-{{x}^{2}} \\right)}{8\\left( {{x}^{2}}-4 \\right)} $<br\/>$ =-\\dfrac{1}{4} $ <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 $A=-\\dfrac{1}{4}$. <\/span><\/span> "}]}],"id_ques":226},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/5.jpg' \/><\/center> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x,$ th\u1ecfa m\u00e3n bi\u1ec3u th\u1ee9c $Q=\\dfrac{{{x}^{2}}-25}{x+\\dfrac{20x+25}{x}}$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng 0.<br\/> \u0110\u00e1p \u00e1n: $x =$ _input_ ","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 $Q.$<br\/> <b> B\u01b0\u1edbc 2:<\/b> Cho bi\u1ec3u th\u1ee9c $Q$ b\u1eb1ng $0$ \u0111\u1ec3 t\u00ecm $x.$ <br\/> <b> B\u01b0\u1edbc 3:<\/b> So s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n \u0111\u1ec3 k\u1ebft lu\u1eadn $x$ t\u00ecm \u0111\u01b0\u1ee3c. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n:<br\/> $x\\ne 0$; $x+\\dfrac{20x+25}{x} \\ne 0$ <br\/> $Q = 0\\Leftrightarrow$ $ \\dfrac{{{x}^{2}}-25}{x+\\dfrac{20x+25}{x}}=0 $ <br\/> $ \\Rightarrow x^2-25=0 $ <br\/> $ \\Leftrightarrow (x+5)(x-5)=0 $ <br\/> $\\Leftrightarrow \\left[ \\begin{aligned} & x-5=0 \\\\ & x+5=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=5\\,\\,\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ & x=-5\\,\\,\\,\\left( \\text{lo\u1ea1i} \\right) \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $5$. <\/span><\/span> "}]}],"id_ques":227},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/4.jpg' \/><\/center> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x,$ th\u1ecfa m\u00e3n bi\u1ec3u th\u1ee9c $M=\\dfrac{7{{x}^{2}}-14x+7}{3x^2+3x}$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng $0.$<br\/> \u0110\u00e1p \u00e1n: $x =$ _input_ ","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 $M.$<br\/> <b> B\u01b0\u1edbc 2:<\/b> Cho bi\u1ec3u th\u1ee9c $M$ b\u1eb1ng $0$ \u0111\u1ec3 t\u00ecm $x.$ <br\/> <b> B\u01b0\u1edbc 3:<\/b> So s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n \u0111\u1ec3 k\u1ebft lu\u1eadn $x$ t\u00ecm \u0111\u01b0\u1ee3c. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n:<br\/> $x\\ne \\{-1;0\\}$ <br\/> $M = 0 \\Leftrightarrow$ $ \\dfrac{7{{x}^{2}}-14x+7}{3x^2+3x}=0 $ <br\/> $ \\Rightarrow 7x^2-14x+7=0 $ <br\/> $ \\Leftrightarrow 7(x^2-2x+1)=0 $ <br\/> $ \\Leftrightarrow 7(x-1)^2=0$ <br\/> $ \\Leftrightarrow x-1=0 $ <br\/> $ \\Leftrightarrow x=1 \\, ( \\text{th\u1ecfa m\u00e3n}) $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$. <\/span><\/span> "}]}],"id_ques":228},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/3.jpg' \/><\/center> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x,$ th\u1ecfa m\u00e3n bi\u1ec3u th\u1ee9c $Q=\\dfrac{{{x}^{4}}+{{x}^{3}}+x+1}{{{x}^{4}}-{{x}^{3}}+2{{x}^{2}}-x+1}$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng $0.$<br\/> \u0110\u00e1p \u00e1n: $x =$ _input_ ","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 $Q.$<br\/> <b> B\u01b0\u1edbc 2:<\/b> Cho bi\u1ec3u th\u1ee9c $Q$ b\u1eb1ng $0$ \u0111\u1ec3 t\u00ecm $x.$ <br\/> <b> B\u01b0\u1edbc 3:<\/b> So s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n \u0111\u1ec3 k\u1ebft lu\u1eadn $x$ t\u00ecm \u0111\u01b0\u1ee3c. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n:<br\/> $x^4-x^3+2x^2-x+1 \\ne 0$ <br\/> $Q = 0\\Leftrightarrow$ $ \\dfrac{{{x}^{4}}+{{x}^{3}}+x+1}{{{x}^{4}}-{{x}^{3}}+2{{x}^{2}}-x+1} =0 $ <br\/> $ \\Rightarrow x^4+x^3+x+1=0 $ <br\/> $ \\Leftrightarrow x^3(x+1)+(x+1)=0 $ <br\/> $ \\Leftrightarrow (x+1)(x^3+1)=0 $ <br\/> $ \\Leftrightarrow (x+1)^2(x^2-x+1)=0 $ <br\/> $\\Leftrightarrow \\left[ \\begin{aligned} & x+1=0 \\\\ & x^2-x+1=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=-1\\,\\,\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ & x^2-x+1 \\ne 0 \\, \\forall x \\,\\, \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$. <\/span><\/span> "}]}],"id_ques":229},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-3"],["20"]]],"list":[{"point":5,"width":50,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/2.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{6-{{x}^{2}}}{3{{x}^{2}}+2x-1}$ t\u1ea1i $x = -3$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":" <span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\left\\{-1;\\dfrac{1}{3}\\right\\}$ <br\/> Thay $x = -3$ v\u00e0o bi\u1ec3u th\u1ee9c, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{6-{{\\left( -3 \\right)}^{2}}}{3.{{\\left( -3 \\right)}^{2}}+2.\\left( -3 \\right)-1}$$=\\dfrac{6-9}{27-6-1}=\\dfrac{-3}{20}$ <\/span> "}]}],"id_ques":230},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["10"],["9"]]],"list":[{"point":5,"width":50,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/16.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\left( \\dfrac{2x+1}{2x-1}-\\dfrac{2x-1}{2x+1} \\right)$$:\\dfrac{4x}{10x-5} $ t\u1ea1i $x = 4$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x = 4$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\neq \\{0;\\pm \\dfrac{1}{2}\\}$ <br\/> Ta c\u00f3: <br\/> $ \\left( \\dfrac{2x+1}{2x-1}-\\dfrac{2x-1}{2x+1} \\right)$$:\\dfrac{4x}{10x-5} $<br\/>$ =\\left[ \\dfrac{{{\\left( 2x+1 \\right)}^{2}}}{\\left( 2x+1 \\right)\\left( 2x-1 \\right)}-\\dfrac{{{\\left( 2x-1 \\right)}^{2}}}{\\left( 2x+1 \\right)\\left( 2x-1 \\right)} \\right]$$:\\dfrac{4x}{5\\left( 2x-1 \\right)} $<br\/>$ =\\dfrac{{{\\left( 2x+1 \\right)}^{2}}-{{\\left( 2x-1 \\right)}^{2}}}{\\left( 2x+1 \\right)\\left( 2x-1 \\right)}$$\\cdot \\dfrac{5\\left( 2x-1 \\right)}{4x} $<br\/>$ =\\dfrac{\\left( 2x+1+2x-1 \\right)\\left( 2x+1-2x+1 \\right)}{\\left( 2x+1 \\right)\\left( 2x-1 \\right)}$$\\cdot \\dfrac{5\\left( 2x-1 \\right)}{4x} $<br\/>$ =\\dfrac{4x.2.5\\left( 2x-1 \\right)}{\\left( 2x+1 \\right)\\left( 2x-1 \\right)4x} $<br\/>$ =\\dfrac{10}{2x+1} $ <br\/> Thay $x = 4$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{10}{2x+1} = \\dfrac{10}{2.4+1}=\\dfrac{10}{9}$ <\/span> "}]}],"id_ques":231},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["5"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/13.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\left( \\dfrac{1}{{{x}^{2}}+x}-\\dfrac{2-x}{x+1} \\right)$$:\\left( \\dfrac{1}{x}+x-2 \\right)$ t\u1ea1i $x = 4$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","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 $x = 4$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><span class='basic_left'> <br\/> \u0110i\u1ec1u ki\u1ec7n: $x \\neq \\{0;\\pm 1\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\left( \\dfrac{1}{{{x}^{2}}+x}-\\dfrac{2-x}{x+1} \\right):\\left( \\dfrac{1}{x}+x-2 \\right) \\\\ & =\\left[ \\dfrac{1}{x\\left( x+1 \\right)}-\\dfrac{2-x}{x+1} \\right]:\\left[ \\dfrac{1}{x}+\\dfrac{{{x}^{2}}}{x}-\\dfrac{2x}{x} \\right] \\\\ & =\\left[ \\dfrac{1-\\left( 2-x \\right)x}{x\\left( x+1 \\right)} \\right]:\\dfrac{1+{{x}^{2}}-2x}{x} \\\\ & =\\dfrac{1-2x+{{x}^{2}}}{x\\left( x+1 \\right)}:\\dfrac{{{\\left( x-1 \\right)}^{2}}}{x} \\\\ & =\\dfrac{{{\\left( x-1 \\right)}^{2}}}{x\\left( x+1 \\right)}\\cdot \\dfrac{x}{{{\\left( x-1 \\right)}^{2}}} \\\\ & =\\dfrac{1}{x+1} \\\\ \\end{align}$ <br\/> Thay $x = 4$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{1}{x+1}=\\dfrac{1}{4+1}=\\dfrac{1}{5}$<\/span> "}]}],"id_ques":232},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/12.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{{{x}^{2}}+3x+2}{{{x}^{3}}+2{{x}^{2}}-x-2}$ l\u00e0: ","select":["A. $\\dfrac{x-1}{x+1}$ ","B. $\\dfrac{x+1}{x-1}$","C. $\\dfrac{1}{x+1}$","D. $\\dfrac{1}{x-1}$"],"hint":" Ph\u00e2n t\u00edch t\u1eed th\u1ee9c v\u00e0 m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i r\u00fat g\u1ecdn. ","explain":"<span class='basic_left'><br\/> \u0110i\u1ec1u ki\u1ec7n: $x \\neq \\{-2;\\pm 1\\}$ <br\/> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{{{x}^{2}}+3x+2}{{{x}^{3}}+2{{x}^{2}}-x-2} \\\\ & =\\dfrac{{{x}^{2}}+x+2x+2}{{{x}^{2}}\\left( x+2 \\right)-\\left( x+2 \\right)} \\\\ & =\\dfrac{x\\left( x+1 \\right)+2\\left( x+1 \\right)}{\\left( x+2 \\right)\\left( {{x}^{2}}-1 \\right)} \\\\ & =\\dfrac{\\left( x+1 \\right)\\left( x+2 \\right)}{\\left( x+2 \\right)\\left( x+1 \\right)\\left( x-1 \\right)} \\\\ & =\\dfrac{1}{x-1} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span><\/span> ","column":2}]}],"id_ques":233},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/11.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{2-\\dfrac{a}{x}}{2+\\dfrac{a}{x}}$ l\u00e0: ","select":["A. $\\dfrac{2-a}{2+a}$ ","B. $-1$","C. $\\dfrac{2x-a}{2x+a}$","D. $\\dfrac{x-a}{x+a}$"],"explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne 0; 2+\\dfrac{a}{x}\\ne 0$ <br\/> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{2-\\dfrac{a}{x}}{2+\\dfrac{a}{x}}=\\dfrac{\\dfrac{2x-a}{x}}{\\dfrac{2x+a}{x}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\dfrac{2x-a}{x}\\cdot \\dfrac{x}{2x+a} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\dfrac{2x-a}{2x+a} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span><\/span> ","column":2}]}],"id_ques":234},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-1"],["-2"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $C=\\dfrac{1}{\\left( x+1 \\right)\\left( x+2 \\right)}+\\dfrac{1}{x+1}-\\dfrac{1}{x+2}$ <br\/> <br\/> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $C$ l\u00e0 $\\left\\{ \\begin{aligned} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $C$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & (x+1)(x+2)\\ne 0 \\\\ & x+1\\ne 0 \\\\ & x+2\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & x+1\\ne 0 \\\\ & x+2\\ne 0 \\\\ \\end{aligned} \\right.$ $\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne -1 \\\\ & x\\ne -2 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2; -1.$ <\/span><\/span> "}]}],"id_ques":235},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["0"],["1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\dfrac{a+1}{{{a}^{3}}-1}\\cdot \\dfrac{{{a}^{2}}+1}{a}+\\dfrac{2a+2}{{{a}^{2}}+a+1}$ <br\/> <br\/> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a B l\u00e0 $\\left\\{ \\begin{aligned} & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $B$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & a^3-1\\ne 0 \\\\ & a\\ne 0 \\\\ & a^2+a+1\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & (a-1)(a^2+a+1)\\ne 0 \\\\ & a\\ne 0 \\\\ & a^2+a+1\\ne 0 \\,\\, \\forall a \\\\ \\end{aligned} \\right.$ $\\Rightarrow \\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & a\\ne 1 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0; 1.$ <\/span><\/span> "}]}],"id_ques":236},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\left\\{ \\begin{aligned} & x\\ne \\frac{1}{2} \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$","B. $\\left\\{ \\begin{aligned} & x\\ne 1 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$","C. $\\left\\{ \\begin{aligned} & x\\ne 2 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/9.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\left( 2x+1-\\dfrac{1}{1-2x} \\right):\\left( 2x-\\dfrac{4{{x}^{2}}}{2x-1} \\right)$ <br\/> <br\/> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $A$ ?<\/span>","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $A$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned}& 1-2x\\ne 0 \\\\ & 2x-\\frac{4{{x}^{2}}}{2x-1}\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned}& x\\ne \\frac{1}{2} \\\\ & \\frac{-2x}{2x-1}\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned}& x\\ne \\frac{1}{2} \\\\ & -2x\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne \\frac{1}{2} \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$<\/span> "}]}],"id_ques":237},{"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":[[["6"],["x-y"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/5.jpg' \/><\/center> <span class='basic_left'> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/> $\\dfrac{9x-9y}{\\left( 2{{x}^{2}}+2xy+2{{y}^{2}} \\right)}\\cdot \\dfrac{4{{x}^{3}}-4{{y}^{3}}}{3x-3y}$$=(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\neq y$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{9x-9y}{\\left( 2{{x}^{2}}+2xy+2{{y}^{2}} \\right)}\\cdot \\dfrac{4{{x}^{3}}-4{{y}^{3}}}{3x-3y} \\\\ & =\\dfrac{9\\left( x-y \\right).4\\left( {{x}^{3}}-{{y}^{3}} \\right)}{2\\left( {{x}^{2}}+xy+{{y}^{2}} \\right).3\\left( x-y \\right)} \\\\ & =\\dfrac{3.2\\left( x-y \\right)\\left( {{x}^{3}}-{{y}^{3}} \\right)}{\\left( {{x}^{3}}-{{y}^{3}} \\right)} \\\\ & =6\\left( x-y \\right) \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $6$ v\u00e0 $x-y$. <\/span><\/span> "}]}],"id_ques":238},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/4.jpg' \/><\/center> <span class='basic_left'> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/> $\\left( \\dfrac{x+1}{{{x}^{2}}-2x+1}+\\dfrac{1}{x-1} \\right):\\dfrac{x}{x-1}$$-\\dfrac{2}{x-1}$$=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\neq \\{0;1\\}$ <br\/> Ta c\u00f3: <br\/> $ \\left( \\dfrac{x+1}{{{x}^{2}}-2x+1}+\\dfrac{1}{x-1} \\right):\\dfrac{x}{x-1}$$-\\dfrac{2}{x-1} $<br\/>$ =\\left[ \\dfrac{x+1}{{{\\left( x-1 \\right)}^{2}}}+\\dfrac{x-1}{{{\\left( x-1 \\right)}^{2}}} \\right]:\\dfrac{x}{x-1}$$-\\dfrac{2}{x-1} $<br\/>$ =\\dfrac{x+1+x-1}{{{\\left( x-1 \\right)}^{2}}}:\\dfrac{x}{x-1}$$-\\dfrac{2}{x-1} $<br\/>$ =\\dfrac{2x}{{{\\left( x-1 \\right)}^{2}}}\\cdot \\dfrac{x-1}{x}$$-\\dfrac{2}{x-1} $<br\/>$ =\\dfrac{2}{x-1}-\\dfrac{2}{x-1} $<br\/>$ =0 $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0.$ <\/span><\/span> "}]}],"id_ques":239},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["2x"],["x+y","y+x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv2/img\/3.jpg' \/><\/center> <span class='basic_left'> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/> $\\dfrac{4xy}{{{y}^{2}}-{{x}^{2}}}$$:\\left( \\dfrac{1}{{{y}^{2}}-{{x}^{2}}}+\\dfrac{1}{{{x}^{2}}+2xy+{{y}^{2}}} \\right)$$=(\\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: $x\\neq \\pm y; y \\ne 0$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{4xy}{{{y}^{2}}-{{x}^{2}}}:\\left( \\dfrac{1}{{{y}^{2}}-{{x}^{2}}}+\\dfrac{1}{{{x}^{2}}+2xy+{{y}^{2}}} \\right) \\\\ & =\\dfrac{4xy}{{{y}^{2}}-{{x}^{2}}}:\\left[ \\dfrac{1}{\\left( y+x \\right)\\left( y-x \\right)}+\\dfrac{1}{{{\\left( x+y \\right)}^{2}}} \\right] \\\\ & =\\dfrac{4xy}{{{y}^{2}}-{{x}^{2}}}:\\left[ \\dfrac{x+y}{{{\\left( x+y \\right)}^{2}}\\left( y-x \\right)}+\\dfrac{y-x}{{{\\left( x+y \\right)}^{2}}\\left( y-x \\right)} \\right] \\\\ & =\\dfrac{4xy}{{{y}^{2}}-{{x}^{2}}}:\\dfrac{x+y+y-x}{{{\\left( x+y \\right)}^{2}}\\left( y-x \\right)} \\\\ & =\\dfrac{4xy}{\\left( x+y \\right)\\left( y-x \\right)}:\\dfrac{2y}{{{\\left( x+y \\right)}^{2}}\\left( y-x \\right)} \\\\ & =\\dfrac{4xy}{\\left( x+y \\right)\\left( y-x \\right)}\\cdot \\dfrac{{{\\left( x+y \\right)}^{2}}\\left( y-x \\right)}{2y} \\\\ & =2x\\left( x+y \\right) \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2x$ v\u00e0 $x+y$. <\/span><\/span> "}]}],"id_ques":240}],"lesson":{"save":0,"level":2}}