{"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":[[["a"],["1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/10.jpg' \/><\/center> <span class='basic_left'> Cho bi\u1ec3u th\u1ee9c $ P=\\left( {{a}^{2}}-1 \\right)\\left( \\dfrac{1}{a-1}-\\dfrac{1}{a+1}+1 \\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: $P = (\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})^2+\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $a\\ne \\{-1;1\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & P=\\left( {{a}^{2}}-1 \\right)\\left( \\dfrac{1}{a-1}-\\dfrac{1}{a+1}+1 \\right) \\\\ & =\\left( {{a}^{2}}-1 \\right)\\left( \\dfrac{a+1}{{{a}^{2}}-1}-\\dfrac{a-1}{{{a}^{2}}-1}+\\dfrac{{{a}^{2}}-1}{{{a}^{2}}-1} \\right) \\\\ & =\\left( {{a}^{2}}-1 \\right)\\dfrac{a+1-a+1+{{a}^{2}}-1}{{{a}^{2}}-1} \\\\ & =\\dfrac{\\left( {{a}^{2}}-1 \\right)\\left( {{a}^{2}}+1 \\right)}{\\left( {{a}^{2}}-1 \\right)} \\\\ & ={{a}^{2}}+1 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $a$ v\u00e0 $1.$ <\/span><\/span> "}]}],"id_ques":201},{"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 hai c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ P=\\left( {{a}^{2}}-1 \\right)\\left( \\dfrac{1}{a-1}-\\dfrac{1}{a+1}+1 \\right) $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> Ta kh\u1eb3ng \u0111\u1ecbnh \u0111\u01b0\u1ee3c $P > 0$ v\u1edbi m\u1ecdi $a\\ne \\{-1;1\\}$ <\/span> ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'><span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, ta c\u00f3: $P=a^2+1$ <br\/> Do $a^2\\ge 0 \\Rightarrow a^2+1 \\ge 1\\,\\, \\forall a$ <br\/> V\u1eady $P > 0$ v\u1edbi m\u1ecdi $a\\ne \\{-1;1\\}$ <br\/> <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":202},{"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":[[["1"],["-1"],["0"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ Q=\\left( \\dfrac{a+1}{a-1}-\\dfrac{a-1}{a+1} \\right)$$:\\dfrac{2a}{5a-5} $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $Q$ l\u00e0 $\\left\\{ \\begin{aligned} & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $Q$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & a+1\\ne 0 \\\\ & a-1\\ne 0 \\\\ & a\\ne 0\\\\ & 5a - 5\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & a\\ne 1 \\\\ & a \\ne -1 \\\\ & a\\ne 0 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1; -1$ v\u00e0 $0.$ <\/span><\/span> "}]}],"id_ques":203},{"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":[[["10"],["a+1","1+a"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ Q=\\left( \\dfrac{a+1}{a-1}-\\dfrac{a-1}{a+1} \\right)$$:\\dfrac{2a}{5a-5} $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $Q$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $Q = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $a\\ne \\{-1;0;1\\}$ <br\/> Ta c\u00f3: <br\/> $ Q=\\left( \\dfrac{a+1}{a-1}-\\dfrac{a-1}{a+1} \\right)$$:\\dfrac{2a}{5a-5} $<br\/>$ =\\left[ \\dfrac{{{\\left( a+1 \\right)}^{2}}}{\\left( a+1 \\right)\\left( a-1 \\right)}-\\dfrac{{{\\left( a-1 \\right)}^{2}}}{\\left( a+1 \\right)\\left( a-1 \\right)} \\right]$$:\\dfrac{2a}{5\\left( a-1 \\right)} $<br\/>$ =\\dfrac{{{\\left( a+1 \\right)}^{2}}-{{\\left( a-1 \\right)}^{2}}}{\\left( a+1 \\right)\\left( a-1 \\right)}$$:\\dfrac{2a}{5\\left( a-1 \\right)} $<br\/>$ =\\dfrac{\\left( a+1+a-1 \\right)\\left( a+1-a+1 \\right)}{\\left( a+1 \\right)\\left( a-1 \\right)}$$\\cdot \\dfrac{5\\left( a-1 \\right)}{2a} $<br\/>$ =\\dfrac{2a.2.5\\left( a-1 \\right)}{2a\\left( a+1 \\right)\\left( a-1 \\right)} $<br\/>$ =\\dfrac{10}{a+1} $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $\\dfrac{10}{a+1}$. <\/span><\/span> "}]}],"id_ques":204},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/8.jpg' \/><\/center> <span class='basic_left'> Cho bi\u1ec3u th\u1ee9c $ Q=\\left( \\dfrac{a+1}{a-1}-\\dfrac{a-1}{a+1} \\right)$$:\\dfrac{2a}{5a-5} $ <br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> Ta kh\u1eb3ng \u0111\u1ecbnh \u0111\u01b0\u1ee3c $Q > 0$ v\u1edbi m\u1ecdi $a\\ne \\{-1;0;1\\}$ <\/span> ","select":["\u0110\u00fang","Sai"],"hint":" X\u00e9t d\u1ea5u c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a bi\u1ec3u th\u1ee9c $Q$ \u0111\u00e3 r\u00fat g\u1ecdn t\u1eeb \u0111\u00f3 suy ra gi\u00e1 tr\u1ecb c\u1ee7a $Q.$ ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $Q=\\dfrac{10}{a+1}$ <br\/> + Bi\u1ec3u th\u1ee9c $Q$ c\u00f3 t\u1eed th\u1ee9c l\u00e0 $10 > 0$ <br\/> + Bi\u1ec3u th\u1ee9c $Q$ c\u00f3 m\u1eabu l\u00e0 $a+1$ ch\u01b0a x\u00e1c \u0111\u1ecbnh d\u1ea5u v\u1edbi m\u1ecdi $a\\ne \\{-1;0;1\\}$ <br\/> Do \u0111\u00f3 ch\u01b0a th\u1ec3 kh\u1eb3ng \u0111\u1ecbnh $Q > 0 $ v\u1edbi m\u1ecdi $a\\ne \\{-1;0;1\\}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":206},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/4.jpg' \/><\/center> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x,$ th\u1ecfa m\u00e3n bi\u1ec3u th\u1ee9c $E=\\dfrac{3-3x}{{{\\left( 1+x \\right)}^{2}}}$$:\\dfrac{6{{x}^{2}}-6}{x+2}$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng $0.$ ","select":["A. Kh\u00f4ng t\u1ed3n t\u1ea1i $x$ ","B. $x=\\pm 2$","C. $x=-2$","D. $x= 0 $"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u00f3 ngh\u0129a v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $E.$<br\/> <b> B\u01b0\u1edbc 2:<\/b> Cho bi\u1ec3u th\u1ee9c $E$ \u0111\u00e3 r\u00fat g\u1ecdn 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: $x\\ne \\{-2;-1; 1\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & E=\\dfrac{3-3x}{{{\\left( 1+x \\right)}^{2}}}:\\dfrac{6{{x}^{2}}-6}{x+2} \\\\ & =\\dfrac{3\\left( 1-x \\right)}{{{\\left( 1+x \\right)}^{2}}}:\\dfrac{6\\left( x+1 \\right)\\left( x-1 \\right)}{x+2} \\\\ & =\\dfrac{3\\left( 1-x \\right)}{{{\\left( 1+x \\right)}^{2}}}\\cdot \\dfrac{x+2}{6\\left( x+1 \\right)\\left( x-1 \\right)} \\\\ & =\\dfrac{-\\left( x+2 \\right)}{2{{\\left( x+1 \\right)}^{3}}} \\\\ \\end{align}$ <br\/> $E = 0$ $\\Leftrightarrow \\dfrac{-\\left( x+2 \\right)}{2{{\\left( x+1 \\right)}^{3}}}=0 $ <br\/> $\\Leftrightarrow x+2=0 $ <br\/> $\\Leftrightarrow x=-2 \\, ( \\text{lo\u1ea1i}) $ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":2}]}],"id_ques":207},{"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/lv1/img\/11.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{3{{x}^{2}}-x}{9{{x}^{2}}-6x+1}$ l\u00e0: ","select":["A. $3x-1$ ","B. $3-x$","C. $\\dfrac{1}{3x}$","D. $\\dfrac{x}{3x-1}$"],"hint":"\u0110\u01b0a m\u1eabu th\u1ee9c v\u1ec1 b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t hi\u1ec7u ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{3{{x}^{2}}-x}{9{{x}^{2}}-6x+1} \\\\ & =\\dfrac{x\\left( 3x-1 \\right)}{{{\\left( 3x \\right)}^{2}}-2.3x.1+{{1}^{2}}} \\\\ & =\\dfrac{x\\left( 3x-1 \\right)}{{{\\left( 3x-1 \\right)}^{2}}} \\\\ & =\\dfrac{x}{3x-1} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span><\/span> ","column":2}]}],"id_ques":213},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/10.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{9-12x+4{{x}^{2}}}{2x-3}$ l\u00e0: ","select":["A. $2x-3$ ","B. $2x+3$","C. $(2x-3)^2$","D. $4x-9$"],"explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\dfrac{3}{2}$ <br\/> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{9-12x+4{{x}^{2}}}{2x-3} \\\\ & =\\dfrac{{{3}^{2}}-2.3.2x+{{\\left( 2x \\right)}^{2}}}{2x-3} \\\\ & =\\dfrac{{{\\left( 2x-3 \\right)}^{2}}}{2x-3} \\\\ & =2x-3 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":2}]}],"id_ques":214},{"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/lv1/img\/9.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{1}{{{x}^{2}}+x}-\\dfrac{2}{x+1} \\right)$$:\\dfrac{x}{x+1}$ <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^2+x\\ne 0 \\\\ & x+1\\ne 0 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & x(x+1)\\ne 0 \\\\ & x+1\\ne 0 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$ $\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne -1 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$ v\u00e0 $0.$ <\/span><\/span> "}]}],"id_ques":215},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-1"],["-2"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\dfrac{x}{\\left( x+1 \\right)\\left( x+2 \\right)}-\\dfrac{1}{x+1}-\\dfrac{1}{x+2}$ <br\/> <br\/> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $B$ l\u00e0: $\\left\\{ \\begin{aligned} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $B$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x+2\\ne 0 \\\\ & x+1\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne -2 \\\\ & 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 $-2$ v\u00e0 $-1.$ <\/span><\/span> "}]}],"id_ques":216},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","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/lv1/img\/5.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{3}{{{\\left( x-2 \\right)}^{2}}}+\\dfrac{1}{x+2}+\\dfrac{x}{x-2}$ <br\/> <br\/> \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} & x-2\\ne 0 \\\\ & x+2\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 2 \\\\ & x\\ne -2 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2$ v\u00e0 $2.$ <\/span><\/span> "}]}],"id_ques":217},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["9a"],["a+2","2+a"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/4.jpg' \/><\/center> <span class='basic_left'> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/> $\\left( \\dfrac{a+6}{3a+9}-\\dfrac{1}{a+3} \\right):\\dfrac{a+2}{27a} $$=\\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: $a\\ne \\{-3;-2;0\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\left( \\dfrac{a+6}{3a+9}-\\dfrac{1}{a+3} \\right):\\dfrac{a+2}{27a} \\\\ & =\\left[ \\dfrac{\\left( a+6 \\right)}{3\\left( a+3 \\right)}-\\dfrac{1}{a+3} \\right]:\\dfrac{a+2}{27a} \\\\ & = \\dfrac{a+6-3}{3\\left( a+3 \\right)} \\cdot \\dfrac{27a}{a+2} \\\\ & =\\dfrac{a+3}{3\\left( a+3 \\right)}\\cdot \\dfrac{27a}{a+2} \\\\ & =\\dfrac{1}{3}\\cdot \\dfrac{27a}{a+2} \\\\ & =\\dfrac{9a}{a+2} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $9a$ v\u00e0 $a+2$. <\/span><\/span> "}]}],"id_ques":218},{"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":[[["1-x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/3.jpg' \/><\/center> <span class='basic_left'> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/> $\\left( \\dfrac{x-3}{x}-\\dfrac{x}{x-3}+\\dfrac{9}{{{x}^{2}}-3x} \\right)$$:\\dfrac{2x-2}{x} $$=\\dfrac{3}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{0;1;3\\}$ <br\/> Ta c\u00f3: <br\/> $\\left( \\dfrac{x-3}{x}-\\dfrac{x}{x-3}+\\dfrac{9}{{{x}^{2}}-3x} \\right)$$:\\dfrac{2x-2}{x} $ <br\/> $ =\\left[ \\dfrac{{{\\left( x-3 \\right)}^{2}}}{x\\left( x-3 \\right)}-\\dfrac{{{x}^{2}}}{x\\left( x-3 \\right)}+\\dfrac{9}{x\\left( x-3 \\right)} \\right]$$:\\dfrac{2\\left( x-1 \\right)}{x} $ <br\/> $ =\\dfrac{{{\\left( x-3 \\right)}^{2}}-{{x}^{2}}+9}{x\\left( x-3 \\right)}$$\\cdot \\dfrac{x}{2\\left( x-1 \\right)} $ <br\/> $ =\\dfrac{{{x}^{2}}-6x+9-{{x}^{2}}+9}{x\\left( x-3 \\right)}$$\\cdot \\dfrac{x}{2\\left( x-1 \\right)} $ <br\/> $=\\dfrac{-6x+18}{x\\left( x-3 \\right)}\\cdot \\dfrac{x}{2\\left( x-1 \\right)} $<br\/>$ =\\dfrac{-6\\left( x-3 \\right)}{x\\left( x-3 \\right)}$$\\cdot \\dfrac{x}{2\\left( x-1 \\right)} $ <br\/> $=\\dfrac{-6x}{2x\\left( x-1 \\right)} $ <br\/> $=\\dfrac{-3}{x-1} $ <br\/> $ =\\dfrac{3}{1-x} $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1-x$. <\/span><\/span> "}]}],"id_ques":219},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai16/lv1/img\/2.jpg' \/><\/center> <span class='basic_left'> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/> $-\\dfrac{5{{x}^{2}}-10x}{3{{x}^{2}}}$$\\cdot \\dfrac{15{{x}^{3}}}{8-{{x}^{3}}} =\\dfrac{(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})^2}{x^2+2x+4}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{0;2\\}$ <br\/> Ta c\u00f3: <br\/> $-\\dfrac{5{{x}^{2}}-10x}{3{{x}^{2}}}$$\\cdot \\dfrac{15{{x}^{3}}}{8-{{x}^{3}}} $ <br\/> $ =-\\dfrac{5x\\left( x-2 \\right)}{3{{x}^{2}}}$$\\cdot \\dfrac{15{{x}^{3}}}{\\left( 2-x \\right)\\left( 4+2x+{{x}^{2}} \\right)} $ <br\/> $ =\\dfrac{5\\left( 2-x \\right)}{3x}$$\\cdot \\dfrac{15{{x}^{3}}}{\\left( 2-x \\right)\\left( 4+2x+{{x}^{2}} \\right)} $ <br\/> $ =\\dfrac{5.5{{x}^{2}}}{4+2x+{{x}^{2}}} $<br\/>$ =\\dfrac{{{(5x)}^{2}}}{{{x}^{2}}+2x+4} $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $5x$. <\/span><\/span> "}]}],"id_ques":220}],"lesson":{"save":0,"level":1}}