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{"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}}

Điểm của bạn.

Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên + 5 điểm
Trong khoảng 1 phút -> 2 phút + 4 điểm
Trong khoảng 2 phút -> 3 phút + 3 điểm
Trong khoảng 3 phút -> 4 phút + 2 điểm
Trong khoảng 4 phút -> 5 phút + 1 điểm

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

Điểm của bạn.

Bấm vào đây nếu phát hiện có lỗi hoặc muốn gửi góp ý