{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/12.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ A=\\dfrac{y}{3-y}$$+\\dfrac{{{y}^{2}}+3y}{2y+3}$$\\cdot \\left( \\dfrac{y+3}{{{y}^{2}}-3y}-\\dfrac{y}{{{y}^{2}}-9} \\right) $<br\/> <br\/> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $A$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ebfn $y$ <\/span> ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $y\\ne \\{-3; -\\dfrac{3}{2}; 0;3\\}$ <br\/> Ta c\u00f3: <br\/> $ A=\\dfrac{y}{3-y}$$+\\dfrac{{{y}^{2}}+3y}{2y+3}$$\\cdot \\left( \\dfrac{y+3}{{{y}^{2}}-3y}-\\dfrac{y}{{{y}^{2}}-9} \\right) $<br\/>$=\\dfrac{y}{3-y}$$+\\dfrac{y\\left( y+3 \\right)}{2y+3}$$\\cdot \\left[ \\dfrac{y+3}{y\\left( y-3 \\right)}-\\dfrac{y}{\\left( y+3 \\right)\\left( y-3 \\right)} \\right] $<br\/>$ =\\dfrac{y}{3-y}$$+\\dfrac{y\\left( y+3 \\right)}{2y+3}$$\\cdot \\dfrac{{{\\left( y+3 \\right)}^{2}}-{{y}^{2}}}{y\\left( y+3 \\right)\\left( y-3 \\right)} $<br\/>$ =\\dfrac{y}{3-y}$$+\\dfrac{y\\left( y+3 \\right)}{2y+3}$$\\cdot \\dfrac{\\left( y+3-y \\right)\\left( y+3+y \\right)}{y\\left( y+3 \\right)\\left( y-3 \\right)} $<br\/>$ =\\dfrac{y}{3-y}$$+\\dfrac{y\\left( y+3 \\right)}{2y+3}\\cdot \\dfrac{3\\left( 2y+3 \\right)}{y\\left( y+3 \\right)\\left( y-3 \\right)} $<br\/>$ =\\dfrac{y}{3-y}+\\dfrac{3}{y-3} $<br\/>$ =\\dfrac{-y+3}{y-3} $<br\/>$ =-1 $ <br\/> V\u1eady gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $A$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ebfn $y.$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang. <\/span><\/span> ","column":2}]}],"id_ques":171},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["1","0","-1"],["0","1","-1"],["-1","0","1"],["-1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/11.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ P=\\dfrac{1-{{a}^{2}}}{1+b}\\cdot \\dfrac{1-{{b}^{2}}}{{{a}^{2}}+a}$$\\cdot \\left( 1+\\dfrac{a}{1-a} \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $P$ l\u00e0: $\\left\\{ \\begin{aligned} & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & b\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $P$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & a^2+a\\ne 0 \\\\ & 1-a\\ne 0 \\\\ & 1+b\\ne 0 \\\\\\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & a\\ne \\pm 1 \\\\ & b\\ne -1 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1; 1; 0$ v\u00e0 $-1.$ <\/span><\/span> "}]}],"id_ques":172},{"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":[[["1-b","-b+1"],["a"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/11.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ P=\\dfrac{1-{{a}^{2}}}{1+b}\\cdot \\dfrac{1-{{b}^{2}}}{{{a}^{2}}+a}$$\\cdot \\left( 1+\\dfrac{a}{1-a} \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $P$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $P = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $a\\ne \\{-1;0; 1\\}; b\\ne -1$ <br\/> Ta c\u00f3: <br\/> $ P=\\dfrac{1-{{a}^{2}}}{1+b}\\cdot \\dfrac{1-{{b}^{2}}}{{{a}^{2}}+a}$$\\cdot \\left( 1+\\dfrac{a}{1-a} \\right) $<br\/>$=\\dfrac{\\left( 1+a \\right)\\left( 1-a \\right)}{1+b}$$\\cdot \\dfrac{\\left( 1+b \\right)\\left( 1-b \\right)}{a\\left( a+1 \\right)}$$\\cdot \\left( \\dfrac{1-a}{1-a}+\\dfrac{a}{1-a} \\right) $<br\/>$=\\dfrac{\\left( 1-a \\right)\\left( 1-b \\right)}{a}$$\\cdot \\dfrac{1-a+a}{1-a} $<br\/>$ =\\dfrac{\\left( 1-a \\right)\\left( 1-b \\right)}{a}$$\\cdot \\dfrac{1}{1-a} $<br\/>$ =\\dfrac{1-b}{a} $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1-b$ v\u00e0 $a$. <\/span><\/span> "}]}],"id_ques":173},{"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":[[["-2016"],["2017"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/11.jpg' \/><\/center>Cho bi\u1ec3u th\u1ee9c $ P=\\dfrac{1-{{a}^{2}}}{1+b}\\cdot \\dfrac{1-{{b}^{2}}}{{{a}^{2}}+a}$$\\cdot \\left( 1+\\dfrac{a}{1-a} \\right) $<br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> V\u1edbi $a=b=2017$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $P$ l\u00e0<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $P=\\dfrac{1-b}{a}$ <br\/> Thay $a=b=2017$ v\u00e0o bi\u1ec3u th\u1ee9c $P$ \u0111\u00e3 r\u00fat g\u1ecdn tr\u00ean, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{1-b}{a}=\\dfrac{1-2017}{2017}$$=\\dfrac{-2016}{2017}$<\/span> "}]}],"id_ques":174},{"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":[[["1"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/11.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ P=\\dfrac{1-{{a}^{2}}}{1+b}\\cdot \\dfrac{1-{{b}^{2}}}{{{a}^{2}}+a}$$\\cdot \\left( 1+\\dfrac{a}{1-a} \\right) $<br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> \u0110\u1ec3 $P=0$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $b$ l\u00e0 _input_ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $P=\\dfrac{1-b}{a}$ <br\/> Khi \u0111\u00f3:<br\/> $P = 0 \\Leftrightarrow \\dfrac{1-b}{a}=0 $ <br\/> $ \\Leftrightarrow 1-b=0 $ <br\/> $ \\Leftrightarrow b=1 (\\text{th\u1ecfa m\u00e3n}) $<\/span> "}]}],"id_ques":175},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["x"],["y"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/9.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ E=\\left( x-\\dfrac{4xy}{x+y}+y \\right)$$:\\left( \\dfrac{x}{x+y}-\\dfrac{y}{y-x}-\\dfrac{2xy}{{{x}^{2}}-{{y}^{2}}} \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $E$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $E = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}-\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\pm y \\ne 0$ <br\/> Ta c\u00f3: <br\/> $ E=\\left( x-\\dfrac{4xy}{x+y}+y \\right)$$:\\left( \\dfrac{x}{x+y}-\\dfrac{y}{y-x}-\\dfrac{2xy}{{{x}^{2}}-{{y}^{2}}} \\right) $<br\/>$=\\left[ \\dfrac{x\\left( x+y \\right)}{x+y}-\\dfrac{4xy}{x+y}+\\dfrac{y\\left( x+y \\right)}{x+y} \\right]$$:\\left[ \\dfrac{x\\left( x-y \\right)}{{{x}^{2}}-{{y}^{2}}}+\\dfrac{y\\left( x+y \\right)}{{{x}^{2}}-{{y}^{2}}}-\\dfrac{2xy}{{{x}^{2}}-{{y}^{2}}} \\right] $<br\/>$ =\\dfrac{x\\left( x+y \\right)-4xy+y\\left( x+y \\right)}{x+y}$$:\\dfrac{x\\left( x-y \\right)+y\\left( x+y \\right)-2xy}{{{x}^{2}}-{{y}^{2}}} $<br\/>$=\\dfrac{{{x}^{2}}+xy-4xy+xy+{{y}^{2}}}{x+y}$$:\\dfrac{{{x}^{2}}-xy+xy+{{y}^{2}}-2xy}{\\left( x+y \\right)\\left( x-y \\right)} $<br\/>$ =\\dfrac{{{x}^{2}}-2xy+{{y}^{2}}}{x+y}$$:\\dfrac{{{x}^{2}}-2xy+{{y}^{2}}}{\\left( x+y \\right)\\left( x-y \\right)} $<br\/>$=\\dfrac{{{\\left( x-y \\right)}^{2}}}{x+y}\\cdot \\dfrac{\\left( x+y \\right)\\left( x-y \\right)}{{{\\left( x-y \\right)}^{2}}} $<br\/>$=x-y $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $x$ v\u00e0 $y.$ <\/span><\/span> "}]}],"id_ques":176},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["12"]]],"list":[{"point":5,"width":40,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/9.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ E=\\left( x-\\dfrac{4xy}{x+y}+y \\right)$$:\\left( \\dfrac{x}{x+y}-\\dfrac{y}{y-x}-\\dfrac{2xy}{{{x}^{2}}-{{y}^{2}}} \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> Cho $E=2$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c <br\/> $M=x^2(x+1)-y^2(y-1)$$-3xy(x-y+1)+xy$ l\u00e0 _input_ <\/span> ","hint":" Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $M$ v\u1ec1 k\u1ebft qu\u1ea3 ch\u1ec9 ch\u1ee9a $x-y$ ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, ta c\u00f3: $E=x-y \\Rightarrow x - y = 2$ <br\/> Ta bi\u1ebfn \u0111\u1ed5i $M=x^2(x+1)-y^2(y-1)$$-3xy(x-y+1)+xy$ nh\u01b0 sau:<br\/> $M={{x}^{2}}(x+1)-{{y}^{2}}(y-1)$$-3xy(x-y+1)+xy $<br\/>$ ={{x}^{3}}+{{x}^{2}}-{{y}^{3}}+{{y}^{2}}$$-3{{x}^{2}}y+3x{{y}^{2}}-3xy+xy $<br\/>$ =\\left( {{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}} \\right)$$+\\left( {{x}^{2}}-2xy+{{y}^{2}} \\right) $<br\/>$ ={{\\left( x-y \\right)}^{3}}+{{\\left( x-y \\right)}^{2}} $<br\/> Thay $x-y=2$ v\u00e0o $M$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\begin{align} & M={{\\left( x-y \\right)}^{3}}+{{\\left( x-y \\right)}^{2}} \\\\ & ={{2}^{3}}+{{2}^{2}} \\\\ & =8+4 \\\\ & =12 \\\\ \\end{align}$<\/span> "}]}],"id_ques":177},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["x"],["1"],["2"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/5.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $Q=\\left[ \\dfrac{{{\\left( x-1 \\right)}^{2}}}{3x+{{\\left( x-1 \\right)}^{2}}}-\\dfrac{1-2{{x}^{2}}+4x}{{{x}^{3}}-1}-\\dfrac{1}{1-x} \\right]$$:\\dfrac{2x}{{{x}^{3}}+x} $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $Q$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $Q = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^2+\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{0;1\\}$ <br\/> Ta c\u00f3: <br\/> $Q=\\left[ \\dfrac{{{\\left( x-1 \\right)}^{2}}}{3x+{{\\left( x-1 \\right)}^{2}}}-\\dfrac{1-2{{x}^{2}}+4x}{{{x}^{3}}-1}-\\dfrac{1}{1-x} \\right]$$:\\dfrac{2x}{{{x}^{3}}+x} $<br\/>$=\\left[ \\dfrac{{{\\left( x-1 \\right)}^{2}}}{3x+{{x}^{2}}-2x+1}-\\dfrac{1-2{{x}^{2}}+4x}{{{x}^{3}}-1}-\\dfrac{1}{1-x} \\right]$$:\\dfrac{2x}{{{x}^{3}}+x} $<br\/>$=\\left[ \\dfrac{{{\\left( x-1 \\right)}^{2}}}{{{x}^{2}}+x+1}-\\dfrac{1-2{{x}^{2}}+4x}{{{x}^{3}}-1}-\\dfrac{1}{1-x} \\right]$$:\\dfrac{2x}{{{x}^{3}}+x} $<br\/>$ = \\dfrac{{{\\left( x-1 \\right)}^{3}}-1+2{{x}^{2}}-4x+{{x}^{2}}+x+1}{{{x}^{3}}-1}$$:\\dfrac{2x}{{{x}^{3}}+x} $<br\/>$ =\\dfrac{{{x}^{3}}-3{{x}^{2}}+3x-1+3{{x}^{2}}-3x}{{{x}^{3}}-1}$$:\\dfrac{2x}{{{x}^{3}}+x} $<br\/>$ =\\dfrac{{{x}^{3}}-1}{{{x}^{3}}-1}\\cdot \\dfrac{{{x}^{3}}+x}{2x} $<br\/>$=\\dfrac{x\\left( {{x}^{2}}+1 \\right)}{2x} $<br\/>$ =\\dfrac{{{x}^{2}}+1}{2}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $x ; 1$ v\u00e0 $2$. <\/span><\/span> "}]}],"id_ques":178},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\left[\\begin{array}{l}{x= 2} \\\\ {x = \\dfrac{-3}{2}}\\end{array}\\right.$","B. $\\left[\\begin{array}{l}{x= 2} \\\\ {x = \\dfrac{3}{2}}\\end{array}\\right.$","C. $\\left[\\begin{array}{l}{x= -2} \\\\ {x = \\dfrac{-3}{2}}\\end{array}\\right.$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/5.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $Q=\\left[ \\dfrac{{{\\left( x-1 \\right)}^{2}}}{3x+{{\\left( x-1 \\right)}^{2}}}-\\dfrac{1-2{{x}^{2}}+4x}{{{x}^{3}}-1}-\\dfrac{1}{1-x} \\right]$$:\\dfrac{2x}{{{x}^{3}}+x} $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> \u0110\u1ec3 $4Q=x+8$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $x$ ?<\/span>","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, ta c\u00f3: $Q = \\dfrac{{{x}^{2}}+1}{2}$ <br\/> X\u00e9t $4Q=x+8$, ta \u0111\u01b0\u1ee3c:<br\/> $4.\\dfrac{{{x}^{2}}+1}{2}=x+8$<br\/>$\\begin{aligned} &\\Leftrightarrow \\dfrac{4\\left( {{x}^{2}}+1 \\right)}{2}-x-8=0 \\\\ &\\Leftrightarrow \\dfrac{4\\left( {{x}^{2}}+1 \\right)-2x-16}{2}=0 \\\\ &\\Leftrightarrow \\dfrac{4{{x}^{2}}+4-2x-16}{2}=0 \\\\ &\\Leftrightarrow \\dfrac{4{{x}^{2}}-2x-12}{2}=0 \\\\ &\\Leftrightarrow \\dfrac{2\\left( 2{{x}^{2}}-x-6 \\right)}{2}=0 \\\\ & \\Leftrightarrow 2{{x}^{2}}-x-6=0 \\\\ & \\Leftrightarrow 2{{x}^{2}}+3x-4x-6=0 \\\\ & \\Leftrightarrow x\\left( 2x+3 \\right)-2\\left( 2x+3 \\right)=0 \\\\ & \\Leftrightarrow \\left( x-2 \\right)\\left( 2x+3 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-2=0 \\\\ & 2x+3=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=2\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ & x=-\\dfrac{3}{2}\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$.<\/span> "}]}],"id_ques":179},{"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":[[["-2"],["2"],["5"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/4.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $N=\\dfrac{2x-10}{{{x}^{2}}-7x+10}$$-\\dfrac{2x}{{{x}^{2}}-4}$$+\\dfrac{1}{2-x} $<br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $N$ l\u00e0: $\\left\\{ \\begin{aligned} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $N$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & {{x}^{2}}-7x+10 \\ne 0 \\\\ & {x}^{2} - 4\\ne 0 \\\\ & 2-x\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow\\left\\{ \\begin{aligned} & (x-2)(x-5) \\ne 0 \\\\ & (x-2)(x+2) \\ne 0 \\\\ & x - 2\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x+2\\ne 0 \\\\ & x-2\\ne 0 \\\\ & x-5\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne -2 \\\\ & x\\ne 2 \\\\ & x\\ne 5 \\\\\\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2; 2$ v\u00e0 $5.$ <\/span><\/span> "}]}],"id_ques":180},{"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+2","2+x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/4.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $N=\\dfrac{2x-10}{{{x}^{2}}-7x+10}$$-\\dfrac{2x}{{{x}^{2}}-4}$$+\\dfrac{1}{2-x} $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $N$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $N = \\dfrac{-1}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $x\\ne \\{-2;2;5\\}$ <br\/> Ta c\u00f3: <br\/> $N=\\dfrac{2x-10}{{{x}^{2}}-7x+10}$$-\\dfrac{2x}{{{x}^{2}}-4}$$+\\dfrac{1}{2-x} $<br\/>$ =\\dfrac{2x-10}{{{x}^{2}}-2x-5x+10}$$-\\dfrac{2x}{{{x}^{2}}-4}+\\dfrac{1}{2-x} $<br\/>$ =\\dfrac{2\\left( x-5 \\right)}{x\\left( x-2 \\right)-5\\left( x-2 \\right)}$$-\\dfrac{2x}{{{x}^{2}}-4}+\\dfrac{1}{2-x} $<br\/>$ =\\dfrac{2\\left( x-5 \\right)}{\\left( x-2 \\right)\\left( x-5 \\right)}$$-\\dfrac{2x}{{{x}^{2}}-4}+\\dfrac{1}{2-x} $<br\/>$ =\\dfrac{2}{x-2}$$-\\dfrac{2x}{\\left( x+2 \\right)\\left( x-2 \\right)}$$-\\dfrac{1}{x-2}$<br\/>$ =\\dfrac{1}{x-2}-\\dfrac{2x}{\\left( x+2 \\right)\\left( x-2 \\right)} $<br\/>$ =\\dfrac{1.\\left( x+2 \\right)-2x}{\\left( x+2 \\right)\\left( x-2 \\right)} $<br\/>$ =\\dfrac{2-x}{\\left( x+2 \\right)\\left( x-2 \\right)} $<br\/>$ =\\dfrac{-1}{x+2} $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $x+2$. <\/span><\/span> "}]}],"id_ques":181},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank_random","correct":[[["-3"],["-1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/4.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $N=\\dfrac{2x-10}{{{x}^{2}}-7x+10}$$-\\dfrac{2x}{{{x}^{2}}-4}$$+\\dfrac{1}{2-x} $<br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> T\u00ecm $x$ nguy\u00ean \u0111\u1ec3 $N\\in \\mathbb{Z}$ <br\/> <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x\\in \\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$<\/span> ","hint":" Ph\u00e2n th\u1ee9c $\\dfrac{a}{A}\\in\\mathbb{Z}$ th\u00ec $A$ thu\u1ed9c \u01b0\u1edbc nguy\u00ean c\u1ee7a $a$ v\u1edbi $a\\in \\mathbb{Z}$ ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $ N =-\\dfrac{1}{x+2} $<br\/> \u0110\u1ec3 $N \\in \\mathbb{Z}$ th\u00ec $x+2\\in \u01af(1)$<br\/> M\u00e0 $\u01af(1)=\\{-1; 1\\}$, do \u0111\u00f3 ta c\u00f3 b\u1ea3ng sau:<br\/> <table> <tr> <th>$x + 2$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <\/tr> <tr> <td>$x$<\/td> <td>$-3$<\/td> <td>$-1$<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-3; -1.$ <\/span><\/span> "}]}],"id_ques":182},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/4.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $N=\\dfrac{2x-10}{{{x}^{2}}-7x+10}$$-\\dfrac{2x}{{{x}^{2}}-4}$$+\\dfrac{1}{2-x} $<br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> \u0110\u1ec3 $N=0$ th\u00ec $x = -2 $ <\/span> ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $N=\\dfrac{-1}{x+2}$ <br\/> $N = 0 \\Leftrightarrow \\dfrac{-1}{x+2} =0$ <br\/> $\\Leftrightarrow -1=0$ (v\u00f4 l\u00fd) <br\/> V\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $x$ \u0111\u1ec3 $N=0$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai. <\/span><\/span> ","column":2}]}],"id_ques":183},{"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":[[["0","-1"],["-1","0"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":"\\frac","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/2.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $Q=\\dfrac{{{x}^{3}}+1}{x}\\cdot \\left( \\dfrac{1}{x+1}+\\dfrac{x-1}{{{x}^{2}}-x+1} \\right)$ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $Q$ 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 $Q$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x\\ne 0\\\\ & x + 1\\ne 0 \\\\ & x^2 - x + 1 \\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & x\\ne -1 \\\\ & x^2 - x + 1\\ne 0 \\,\\, \\forall x \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & x\\ne -1 \\\\ \\end{aligned} \\right.$ <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":184},{"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":[[["2x"],["1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/2.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $Q=\\dfrac{{{x}^{3}}+1}{x}\\cdot \\left( \\dfrac{1}{x+1}+\\dfrac{x-1}{{{x}^{2}}-x+1} \\right)$ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $Q$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $Q = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}-\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $x\\ne \\{-1;0 \\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & Q=\\dfrac{{{x}^{3}}+1}{x}\\cdot \\left( \\dfrac{1}{x+1}+\\dfrac{x-1}{{{x}^{2}}-x+1} \\right) \\\\ & =\\dfrac{{{x}^{3}}+1}{x}\\cdot \\left[ \\dfrac{{{x}^{2}}-x+1}{{{x}^{3}}+1}+\\dfrac{\\left( x-1 \\right)\\left( x+1 \\right)}{{{x}^{3}}+1} \\right] \\\\ & =\\dfrac{{{x}^{3}}+1}{x}\\cdot \\dfrac{{{x}^{2}}-x+1+{{x}^{2}}-1}{{{x}^{3}}+1} \\\\ & =\\dfrac{\\left( {{x}^{3}}+1 \\right)\\left( 2{{x}^{2}}-x \\right)}{x\\left( {{x}^{3}}+1 \\right)} \\\\ & =\\dfrac{x\\left( 2x-1 \\right)}{x} \\\\ & =2x-1 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2x$ v\u00e0 $1.$ <\/span><\/span> "}]}],"id_ques":185},{"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":[[["5"]]],"list":[{"point":5,"width":40,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/2.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $Q=\\dfrac{{{x}^{3}}+1}{x}\\cdot \\left( \\dfrac{1}{x+1}+\\dfrac{x-1}{{{x}^{2}}-x+1} \\right)$<br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> V\u1edbi $x=3$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $Q$ l\u00e0 _input_ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $Q=2x-1$ <br\/> Thay $x=3$ v\u00e0o bi\u1ec3u th\u1ee9c $Q$ \u0111\u00e3 r\u00fat g\u1ecdn tr\u00ean, ta \u0111\u01b0\u1ee3c:<br\/> $Q=2.3-1=5$<\/span> "}]}],"id_ques":186},{"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":[[["7"],["2"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/2.jpg' \/><\/center>Cho bi\u1ec3u th\u1ee9c $Q=\\dfrac{{{x}^{3}}+1}{x}\\cdot \\left( \\dfrac{1}{x+1}+\\dfrac{x-1}{{{x}^{2}}-x+1} \\right)$<br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> \u0110\u1ec3 $Q=6$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $Q=2x-1$ <br\/> Cho $Q=6$, ta \u0111\u01b0\u1ee3c: <br\/> $2x-1=6$ <br\/> $\\Leftrightarrow 2x=7$ <br\/> $ \\Leftrightarrow x=\\dfrac{7}{2}$ (th\u1ecfa m\u00e3n) <\/span>"}]}],"id_ques":187},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-1"],["0"],["3"],["2"]]],"list":[{"point":5,"width":30,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/13.jpg' \/><\/center> <span class='basic_left'>T\u00ecm $x$ nguy\u00ean \u0111\u1ec3 $B=\\dfrac{-2}{1-x} \\in \\mathbb{Z}$ <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> ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x \\ne 1$ <br\/> \u0110\u1ec3 $B=\\dfrac{-2}{1-x}=\\dfrac{2}{x-1} \\in \\mathbb{Z}$ th\u00ec $x-1\\in \u01af(2)$<br\/> M\u00e0 $\u01af(2)=\\{-2;-1; 1; 2\\}$, do \u0111\u00f3 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>$-1$<\/td> <td>$0$<\/td> <td>$2$<\/td> <td>$3$<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1; 0; 2; 3.$ <\/span><\/span> "}]}],"id_ques":188},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-3"],["1"],["3"],["7"]]],"list":[{"point":5,"width":30,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/12.jpg' \/><\/center> <span class='basic_left'>T\u00ecm $x$ nguy\u00ean \u0111\u1ec3 $Q=\\dfrac{5}{2-x} \\in \\mathbb{Z}$ <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> ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n $x \\ne 2$ <br\/> \u0110\u1ec3 $Q=\\dfrac{5}{2-x} \\in \\mathbb{Z}$ th\u00ec $2-x\\in \u01af(5)$<br\/> M\u00e0 $\u01af(5)=\\{-5;-1; 1; 5\\}$, do \u0111\u00f3 ta c\u00f3 b\u1ea3ng sau:<br\/> <table> <tr> <th>$2-x$<\/th> <th>$-5$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <th>$5$<\/th> <\/tr> <tr> <td>$x$<\/td> <td>$7$<\/td> <td>$3$<\/td> <td>$1$<\/td> <td>$-3$<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-3; 1; 3; 7.$ <\/span><\/span> "}]}],"id_ques":189},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-4"],["-2"],["0"],["2"]]],"list":[{"point":5,"width":30,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv2/img\/11.jpg' \/><\/center> <span class='basic_left'>T\u00ecm $x$ nguy\u00ean \u0111\u1ec3 $P=\\dfrac{3}{x+1} \\in \\mathbb{Z}$ <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> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne -1$ <br\/> \u0110\u1ec3 $P=\\dfrac{3}{x+1} \\in \\mathbb{Z}$ th\u00ec $x+1\\in \u01af(3)$<br\/> M\u00e0 $\u01af(3)=\\{-3;3;-1;1\\}$, do \u0111\u00f3 ta c\u00f3 b\u1ea3ng sau:<br\/> <table> <tr> <th>$x+1$<\/th> <th>$-3$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <th>$3$<\/th> <\/tr> <tr> <td>$x$<\/td> <td>$-4$<\/td> <td>$-2$<\/td> <td>$0$<\/td> <td>$2$<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-4; -2; 0$ v\u00e0 $2.$ <\/span><\/span> "}]}],"id_ques":190}],"lesson":{"save":0,"level":2}}