{"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_random","correct":[[["1"],["-1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/11.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{2a+2}{{{a}^{2}}+a+1}:\\dfrac{a+1}{{{a}^{3}}-1}$ <br\/> <b> C\u00e2u 1: <\/b> Bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a khi $\\left\\{ \\begin{align} & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & a\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$<\/span> ","hint":"\u0110\u1ec3 bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a th\u00ec m\u1eabu th\u1ee9c kh\u00e1c $0$ v\u00e0 ph\u00e2n th\u1ee9c chia kh\u00e1c $0$","explain":"<span class='basic_left'> Bi\u1ec3u th\u1ee9c $A=\\dfrac{2a+2}{{{a}^{2}}+a+1}:\\dfrac{a+1}{{{a}^{3}}-1}$ c\u00f3 ngh\u0129a khi <br\/> $\\left\\{ \\begin{aligned} & {{a}^{2}}+a+1\\,\\,\\ne 0 \\\\ & {{a}^{3}}-1\\ne 0 \\\\ & a+1\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow\\left\\{ \\begin{aligned} & {{a}^{2}}+a+1\\,\\,\\ne 0\\,\\,\\,\\forall a \\\\ & {{a}^{3}}\\ne 1 \\\\ & a+1\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & a\\ne 1 \\\\ & 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 $1$ v\u00e0 $-1$. <\/span> <br\/> <b> L\u01b0u \u00fd:<\/b> Trong ph\u00e9p chia $\\dfrac{A}{B}:\\dfrac{C}{D}$ th\u00ec \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u00e9p chia c\u00f3 ngh\u0129a l\u00e0 $B;D\\,\\,\\ne 0;\\,\\,\\dfrac{C}{D}\\,\\,\\ne 0$ .<br\/> Do \u0111\u00f3 \u0111\u1ec3 $\\dfrac{A}{B}:\\dfrac{C}{D}$ c\u00f3 ngh\u0129a, ta c\u00f3 th\u1ec3 vi\u1ebft l\u1ea1i th\u00e0nh: $\\left\\{ \\begin{align} & B\\ne 0 \\\\ & D\\ne 0 \\\\ & C\\ne 0 \\\\ \\end{align} \\right.$ <\/span> "}]}],"id_ques":101},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","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/bai14/lv1/img\/11.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{2a+2}{{{a}^{2}}+a+1}:\\dfrac{a+1}{{{a}^{3}}-1}$<br\/> <b> C\u00e2u 2: <\/b> \u0110\u1ec3 $A=0$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $a$ l\u00e0:<\/span> ","select":["A. $a\\in \\{\\varnothing\\} $ ","B. $a\\in \\{0;1\\}$ ","C. $a=1$","D. $a \\in \\{- 1;0;1\\}$"],"hint":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c A.<br\/> Cho $A=0$ \u0111\u1ec3 t\u00ecm $a.$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $ A=\\dfrac{2a+2}{{{a}^{2}}+a+1}:\\dfrac{a+1}{{{a}^{3}}-1} $<br\/> $ =\\dfrac{2\\left( a+1 \\right)}{{{a}^{2}}+a+1}$$:\\dfrac{a+1}{\\left( a-1 \\right)\\left( {{a}^{2}}+a+1 \\right)} $<br\/> $ =\\dfrac{2\\left( a+1 \\right)}{{{a}^{2}}+a+1}$$\\cdot \\dfrac{\\left( a-1 \\right)\\left( {{a}^{2}}+a+1 \\right)}{a+1} $<br\/> $=\\dfrac{2\\left( a+1 \\right)\\left( a-1 \\right)\\left( {{a}^{2}}+a+1 \\right)}{\\left( {{a}^{2}}+a+1 \\right)\\left( a+1 \\right)} $<br\/> $ =2\\left( a-1 \\right)$ <br\/> $A=0\\Leftrightarrow$ $ 2\\left( a-1 \\right)=0 \\\\ \\Leftrightarrow a-1=0 \\\\ \\Leftrightarrow a=1 $ <br\/> K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh $a\\ne \\{-1;1\\}$ suy ra $a=1$ (lo\u1ea1i) <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":102},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> K\u1ebft qu\u1ea3 ph\u00e9p t\u00ednh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/10.jpg' \/><\/center> $\\dfrac{{{\\left( x+y \\right)}^{2}}{{z}^{2}}}{8{{x}^{2}}{{z}^{3}}}\\cdot \\dfrac{4{{x}^{2}}y}{6{{\\left( x+y \\right)}^{4}}}$$=\\dfrac{y}{6{{\\left( x+y \\right)}^{2}}}$ ","select":["\u0110\u00fang","Sai"],"hint":"Ta th\u1ef1c hi\u1ec7n r\u00fat g\u1ecdn v\u1ebf tr\u00e1i r\u1ed3i so s\u00e1nh k\u1ebft qu\u1ea3 v\u1edbi v\u1ebf ph\u1ea3i.","explain":"<span class='basic_left'> Ta c\u00f3: <br\/>V\u1ebf tr\u00e1i = $\\dfrac{{{\\left( x+y \\right)}^{2}}{{z}^{2}}}{8{{x}^{2}}{{z}^{3}}}\\cdot \\dfrac{4{{x}^{2}}y}{6{{\\left( x+y \\right)}^{4}}}$$=\\dfrac{{{\\left( x+y \\right)}^{2}}{{z}^{2}}.4{{x}^{2}}y}{8{{x}^{2}}{{z}^{3}}.6{{\\left( x+y \\right)}^{4}}}$$=\\dfrac{y}{12z{{\\left( x+y \\right)}^{2}}} \\ne$ V\u1ebf ph\u1ea3i <span><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":103},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> K\u1ebft qu\u1ea3 ph\u00e9p t\u00ednh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/9.jpg' \/><\/center> $\\dfrac{216{{x}^{6}}}{343{{y}^{3}}}:\\dfrac{18{{x}^{8}}}{49{{y}^{4}}}\\cdot \\dfrac{7{{x}^{3}}}{4{{y}^{2}}}$$=\\dfrac{3x}{y}$ ","select":["\u0110\u00fang","Sai"],"hint":"Ta th\u1ef1c hi\u1ec7n r\u00fat g\u1ecdn v\u1ebf tr\u00e1i r\u1ed3i so s\u00e1nh k\u1ebft qu\u1ea3 v\u1edbi v\u1ebf ph\u1ea3i.","explain":"<span class='basic_left'> Ta c\u00f3: <br\/>V\u1ebf tr\u00e1i = $\\dfrac{216{{x}^{6}}}{343{{y}^{3}}}:\\dfrac{18{{x}^{8}}}{49{{y}^{4}}}\\cdot \\dfrac{7{{x}^{3}}}{4{{y}^{2}}}$<br\/>$=\\left( \\dfrac{216{{x}^{6}}}{343{{y}^{3}}}\\cdot \\dfrac{49{{y}^{4}}}{18{{x}^{8}}} \\right)\\cdot \\dfrac{7{{x}^{3}}}{4{{y}^{2}}} $<br\/>$ =\\left( \\dfrac{216.49.{{x}^{6}}{{y}^{4}}}{343{{y}^{3}}.18{{x}^{8}}} \\right)\\cdot \\dfrac{7{{x}^{3}}}{4{{y}^{2}}} $<br\/>$ =\\dfrac{12y}{7{{x}^{2}}}\\cdot \\dfrac{7{{x}^{3}}}{4{{y}^{2}}} $<br\/>$ =\\dfrac{12.7{{x}^{3}}y}{7.4{{x}^{2}}{{y}^{2}}} $<br\/>$ =\\dfrac{3x}{y} =$ V\u1ebf ph\u1ea3i <span><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":104},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> K\u1ebft qu\u1ea3 ph\u00e9p t\u00ednh 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/bai14/lv1/img\/8.jpg' \/><\/center> $\\dfrac{{{a}^{2}}b}{{{a}^{2}}-9{{b}^{2}}}\\cdot \\dfrac{a+3b}{2ab}$$=\\dfrac{a}{2\\left( a-3b \\right)}$ ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> V\u1ebf tr\u00e1i = $\\dfrac{{{a}^{2}}b}{{{a}^{2}}-9{{b}^{2}}}\\cdot \\dfrac{a+3b}{2ab}$<br\/>$=\\dfrac{{{a}^{2}}b}{\\left( a+3b \\right)\\left( a-3b \\right)}\\cdot \\dfrac{a+3b}{2ab} $<br\/>$ =\\dfrac{{{a}^{2}}b\\left( a+3b \\right)}{\\left( a+3b \\right)\\left( a-3b \\right)2ab} $<br\/>$ =\\dfrac{a}{2\\left( a-3b \\right)} =$ V\u1ebf ph\u1ea3i <span><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":105},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/5.jpg' \/><\/center> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{1}{{{x}^{2}}+1}:\\dfrac{{{x}^{2}}+2}{{{x}^{2}}}$ l\u00e0:","select":["A. $x\\ne -1$ ","B. $x > -1$ ","C. $x\\ne 0$","D. $x\\ne \\{\\pm 1;\\pm 2\\}$"],"hint":"\u0110\u1ec3 bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a th\u00ec m\u1eabu th\u1ee9c kh\u00e1c $0$ v\u00e0 ph\u00e2n th\u1ee9c chia kh\u00e1c $0$","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{1}{{{x}^{2}}+1}:\\dfrac{{{x}^{2}}+2}{{{x}^{2}}}$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & {{x}^{2}}+1\\ne 0 \\\\ & {{x}^{2}}\\ne 0 \\\\ & {{x}^{2}}+2\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow\\left\\{ \\begin{aligned} & {{x}^{2}}+1\\ne 0\\,\\,\\forall \\,x \\\\ & x\\ne 0 \\\\ & {{x}^{2}}+2\\ne 0\\,\\,\\forall x \\\\ \\end{aligned} \\right.$$\\Rightarrow x\\ne 0$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><br\/> <b> L\u01b0u \u00fd:<\/b> Trong ph\u00e9p chia $\\dfrac{A}{B}:\\dfrac{C}{D}$ th\u00ec \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u00e9p chia c\u00f3 ngh\u0129a l\u00e0 $B;D\\,\\,\\ne 0;\\,\\,\\dfrac{C}{D}\\,\\,\\ne 0$ .<br\/> Do \u0111\u00f3 \u0111\u1ec3 $\\dfrac{A}{B}:\\dfrac{C}{D}$ c\u00f3 ngh\u0129a, ta c\u00f3 th\u1ec3 vi\u1ebft l\u1ea1i th\u00e0nh: $\\left\\{ \\begin{align} & B\\ne 0 \\\\ & D\\ne 0 \\\\ & C\\ne 0 \\\\ \\end{align} \\right.$ ","column":2}]}],"id_ques":106},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/4.jpg' \/><\/center> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\left( x-1 \\right):\\dfrac{{{x}^{2}}-1}{4{{x}^{2}}-4x+1}$ l\u00e0:","select":["A. $x\\ne \\dfrac{1}{2}$ ","B. $x > \\dfrac{1}{2}$ ","C. $x\\ne \\{-1;1\\}$","D. $x\\ne \\{-1;1;\\dfrac{1}{2}\\}$"],"hint":"\u0110\u1ec3 bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a th\u00ec m\u1eabu th\u1ee9c kh\u00e1c $0$ v\u00e0 ph\u00e2n th\u1ee9c chia kh\u00e1c $0$","explain":" <span class='basic_left'>Ta c\u00f3: $\\left( x-1 \\right):\\dfrac{{{x}^{2}}-1}{4{{x}^{2}}-4x+1}$$=\\left( x-1 \\right):\\dfrac{\\left( x+1 \\right)\\left( x-1 \\right)}{{{\\left( 2x-1 \\right)}^{2}}}$ <br\/> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\left( x-1 \\right):\\dfrac{{{x}^{2}}-1}{4{{x}^{2}}-4x+1}$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x+1\\ne 0 \\\\ & x-1\\ne 0 \\\\ & 2x-1\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne -1 \\\\ & x\\ne 1 \\\\ & x\\ne \\dfrac{1}{2} \\\\ \\end{aligned} \\right.$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><br\/> <b> L\u01b0u \u00fd:<\/b> Trong ph\u00e9p chia $\\dfrac{A}{B}:\\dfrac{C}{D}$ th\u00ec \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u00e9p chia c\u00f3 ngh\u0129a l\u00e0 $B;D\\,\\,\\ne 0;\\,\\,\\dfrac{C}{D}\\,\\,\\ne 0$ .<br\/> Do \u0111\u00f3 \u0111\u1ec3 $\\dfrac{A}{B}:\\dfrac{C}{D}$ c\u00f3 ngh\u0129a, ta c\u00f3 th\u1ec3 vi\u1ebft l\u1ea1i th\u00e0nh: $\\left\\{ \\begin{align} & B\\ne 0 \\\\ & D\\ne 0 \\\\ & C\\ne 0 \\\\ \\end{align} \\right.$ ","column":2}]}],"id_ques":107},{"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/bai14/lv1/img\/3.jpg' \/><\/center> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $1:\\left( 1-\\dfrac{1}{a} \\right)$ l\u00e0:","select":["A. $a\\ne \\{0;1\\}$ ","B. $a\\ne 0$ ","C. $a > 0$","D. $a\\ge 0$"],"explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $1:\\left( 1-\\dfrac{1}{a} \\right)$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & 1-\\dfrac{1}{a}\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & \\dfrac{a-1}{a}\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & a-1\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow a\\,\\,\\ne \\,\\,\\left\\{ 0;1 \\right\\}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><br\/> <b> L\u01b0u \u00fd:<\/b> Trong ph\u00e9p chia $\\dfrac{A}{B}:\\dfrac{C}{D}$ th\u00ec \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u00e9p chia c\u00f3 ngh\u0129a l\u00e0 $B;D\\,\\,\\ne 0;\\,\\,\\dfrac{C}{D}\\,\\,\\ne 0$ .<br\/> Do \u0111\u00f3 \u0111\u1ec3 $\\dfrac{A}{B}:\\dfrac{C}{D}$ c\u00f3 ngh\u0129a, ta c\u00f3 th\u1ec3 vi\u1ebft l\u1ea1i th\u00e0nh: $\\left\\{ \\begin{align} & B\\ne 0 \\\\ & D\\ne 0 \\\\ & C\\ne 0 \\\\ \\end{align} \\right.$","column":2}]}],"id_ques":108},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/2.jpg' \/><\/center> Cho bi\u1ec3u th\u1ee9c $ \\dfrac{ab+{{b}^{2}}}{9}:P=\\dfrac{a\\left( a+b \\right)}{3b}.$ Khi \u0111\u00f3 ph\u00e2n th\u1ee9c $P$ l\u00e0:","select":["A. $\\dfrac{3b}{a^2}$ ","B. $\\dfrac{b^2}{3a}$ ","C. $\\dfrac{b}{a}$","D. $\\dfrac{b}{3a}$"],"hint":"Coi $P$ l\u00e0 s\u1ed1 chia. Mu\u1ed1n t\u00ecm s\u1ed1 chia ta l\u1ea5y s\u1ed1 b\u1ecb chia chia cho th\u01b0\u01a1ng <br\/> Th\u1ef1c hi\u1ec7n ph\u00e9p chia ph\u00e2n th\u1ee9c \u0111\u1ec3 t\u00ecm $P$.","explain":"<span class='basic_left'> $\\begin{align} & \\dfrac{ab+{{b}^{2}}}{9}:P=\\dfrac{a\\left( a+b \\right)}{3b} \\\\ &\\Rightarrow P=\\dfrac{ab+{{b}^{2}}}{9}:\\dfrac{a\\left( a+b \\right)}{3b} \\\\ & =\\dfrac{b\\left( a+b \\right)}{9}:\\dfrac{a\\left( a+b \\right)}{3b} \\\\ & =\\dfrac{b\\left( a+b \\right)}{9}\\cdot \\dfrac{3b}{a\\left( a+b \\right)} \\\\ & =\\dfrac{b\\left( a+b \\right)3b}{9a\\left( a+b \\right)} \\\\ & =\\dfrac{{{b}^{2}}}{3a} \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":109},{"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/bai14/lv1/img\/16.jpg' \/><\/center> Cho bi\u1ec3u th\u1ee9c $ \\left( \\dfrac{x}{x+y}+\\dfrac{y}{x-y} \\right)\\cdot P={{x}^{2}}+{{y}^{2}}.$ Khi \u0111\u00f3 ph\u00e2n th\u1ee9c $P$ l\u00e0:","select":["A. $\\dfrac{1}{x+y}$ ","B. $\\dfrac{1}{(x+y)(x-y)}$ ","C. $x^2-y$","D. $x^2-y^2$"],"hint":"Coi $P$ l\u00e0 th\u1eeba s\u1ed1. Mu\u1ed1n t\u00ecm th\u1eeba s\u1ed1 ta l\u1ea5y t\u00edch chia cho th\u1eeba s\u1ed1 \u0111\u00e3 bi\u1ebft. <br\/> Th\u1ef1c hi\u1ec7n r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ecm $P$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> $P=\\left( {{x}^{2}}+{{y}^{2}} \\right)$$:\\left( \\dfrac{x}{x+y}+\\dfrac{y}{x-y} \\right)$ <br\/> <b> B\u01b0\u1edbc 2:<\/b> R\u00fat g\u1ecdn $\\left( \\dfrac{x}{x+y}+\\dfrac{y}{x-y} \\right)$ <br\/> <b> B\u01b0\u1edbc 3:<\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p chia ph\u00e2n th\u1ee9c \u0111\u1ec3 t\u00ecm $P.$ <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> $\\left( \\dfrac{x}{x+y}+\\dfrac{y}{x-y} \\right)\\cdot P={{x}^{2}}+{{y}^{2}} $<br\/> $\\Rightarrow P=\\left( {{x}^{2}}+{{y}^{2}} \\right)$$:\\left( \\dfrac{x}{x+y}+\\dfrac{y}{x-y} \\right) $<br\/> $ =\\left( {{x}^{2}}+{{y}^{2}} \\right)$$:\\left[ \\dfrac{x\\left( x-y \\right)}{\\left( x+y \\right)\\left( x-y \\right)}+\\dfrac{y\\left( x+y \\right)}{\\left( x+y \\right)\\left( x-y \\right)} \\right] $<br\/> $ =\\left( {{x}^{2}}+{{y}^{2}} \\right)$$:\\left[ \\dfrac{x\\left( x-y \\right)+y\\left( x+y \\right)}{\\left( x+y \\right)\\left( x-y \\right)} \\right] $<br\/> $ =\\left( {{x}^{2}}+{{y}^{2}} \\right)$$:\\left[ \\dfrac{{{x}^{2}}-xy+xy+{{y}^{2}}}{\\left( x+y \\right)\\left( x-y \\right)} \\right] $<br\/> $ =\\left( {{x}^{2}}+{{y}^{2}} \\right)$$:\\dfrac{{{x}^{2}}+{{y}^{2}}}{\\left( x+y \\right)\\left( x-y \\right)} $<br\/> $ =\\dfrac{{{x}^{2}}+{{y}^{2}}}{1}\\cdot \\dfrac{\\left( x+y \\right)\\left( x-y \\right)}{{{x}^{2}}+{{y}^{2}}}$<br\/> $ =\\left( x+y \\right)\\left( x-y \\right) $<br\/> $ ={{x}^{2}}-{{y}^{2}} $ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":110},{"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/bai14/lv1/img\/13.jpg' \/><\/center> Cho bi\u1ec3u th\u1ee9c $\\dfrac{{{a}^{2}}+2a}{a-1}\\cdot P=\\dfrac{{{a}^{2}}-4}{{{a}^{2}}-a}.$ Khi \u0111\u00f3 ph\u00e2n th\u1ee9c $P$ l\u00e0:","select":["A. $\\dfrac{a-2}{a^2}$ ","B. $\\dfrac{a}{a-2}$ ","C. $\\dfrac{a-2}{a}$","D. $\\dfrac{1}{a}$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> $P=\\dfrac{{{a}^{2}}-4}{{{a}^{2}}-a}:\\dfrac{{{a}^{2}}+2a}{a-1}$ <br\/> <b> B\u01b0\u1edbc 2:<\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p chia ph\u00e2n th\u1ee9c \u0111\u1ec3 t\u00ecm P. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> $\\begin{align} & \\dfrac{{{a}^{2}}+2a}{a-1}\\cdot P=\\dfrac{{{a}^{2}}-4}{{{a}^{2}}-a} \\\\ &\\Rightarrow P=\\dfrac{{{a}^{2}}-4}{{{a}^{2}}-a}:\\dfrac{{{a}^{2}}+2a}{a-1} \\\\ & =\\dfrac{\\left( a+2 \\right)\\left( a-2 \\right)}{a\\left( a-1 \\right)}:\\dfrac{a\\left( a+2 \\right)}{a-1} \\\\ & =\\dfrac{\\left( a+2 \\right)\\left( a-2 \\right)}{a\\left( a-1 \\right)}\\cdot \\dfrac{a-1}{a\\left( a+2 \\right)} \\\\ & =\\dfrac{\\left( a+2 \\right)\\left( a-2 \\right)\\left( a-1 \\right)}{a\\left( a-1 \\right)a\\left( a+2 \\right)} \\\\ & =\\dfrac{a-2}{{{a}^{2}}} \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":111},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["25"],["19"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/12.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{{{x}^{4}}-{{y}^{4}}}{{{x}^{3}}-{{y}^{3}}}:\\dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}$ t\u1ea1i $x = 3; y = 2$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau \u0111\u00f3 thay $x = 3; y = 2$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $ \\dfrac{{{x}^{4}}-{{y}^{4}}}{{{x}^{3}}-{{y}^{3}}}:\\dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}$<br\/> $=\\dfrac{\\left( {{x}^{2}}+{{y}^{2}} \\right)\\left( {{x}^{2}}-{{y}^{2}} \\right)}{\\left( x-y \\right)\\left( {{x}^{2}}+xy+{{y}^{2}} \\right)}$$:\\dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}} $<br\/> $ =\\dfrac{\\left( {{x}^{2}}+{{y}^{2}} \\right)\\left( {{x}^{2}}-{{y}^{2}} \\right)}{\\left( x-y \\right)\\left( {{x}^{2}}+xy+{{y}^{2}} \\right)}$$\\cdot \\dfrac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}} $<br\/> $ =\\dfrac{\\left( {{x}^{2}}+{{y}^{2}} \\right)\\left( {{x}^{2}}-{{y}^{2}} \\right)\\left( {{x}^{2}}-{{y}^{2}} \\right)}{\\left( x-y \\right)\\left( {{x}^{2}}+xy+{{y}^{2}} \\right)\\left( {{x}^{2}}+{{y}^{2}} \\right)} $<br\/> $ =\\dfrac{{{\\left( {{x}^{2}}-{{y}^{2}} \\right)}^{2}}}{\\left( x-y \\right)\\left( {{x}^{2}}+xy+{{y}^{2}} \\right)} $<br\/> $ =\\dfrac{{{\\left( x-y \\right)}^{2}}{{\\left( x+y \\right)}^{2}}}{\\left( x-y \\right)\\left( {{x}^{2}}+xy+{{y}^{2}} \\right)} $<br\/> $ =\\dfrac{\\left( x-y \\right){{\\left( x+y \\right)}^{2}}}{{{x}^{2}}+xy+{{y}^{2}}}$ <br\/> Thay $x = 3; y = 2$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{\\left( x-y \\right){{\\left( x+y \\right)}^{2}}}{{{x}^{2}}+xy+{{y}^{2}}}$$=\\dfrac{\\left( 3-2 \\right){{\\left( 3+2 \\right)}^{2}}}{{{3}^{2}}+3.2+{{2}^{2}}}=\\dfrac{25}{19}$<\/span> "}]}],"id_ques":112},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/11.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{{{x}^{2}}+{{y}^{2}}}{x+y}\\cdot \\dfrac{{{\\left( x-y \\right)}^{2}}}{{{x}^{2}}}$$-\\dfrac{{{y}^{2}}}{x+y}\\cdot \\dfrac{{{\\left( x-y \\right)}^{2}}}{{{x}^{2}}}$ t\u1ea1i $x = y = 10$ l\u00e0 _input_ ","hint":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau \u0111\u00f3 thay $x = y = 10$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\dfrac{{{x}^{2}}+{{y}^{2}}}{x+y}\\cdot \\dfrac{{{\\left( x-y \\right)}^{2}}}{{{x}^{2}}}$$-\\dfrac{{{y}^{2}}}{x+y}\\cdot \\dfrac{{{\\left( x-y \\right)}^{2}}}{{{x}^{2}}} $<br\/> $=\\dfrac{\\left( {{x}^{2}}+{{y}^{2}} \\right){{\\left( x-y \\right)}^{2}}}{\\left( x+y \\right){{x}^{2}}}$$-\\dfrac{{{y}^{2}}{{\\left( x-y \\right)}^{2}}}{\\left( x+y \\right){{x}^{2}}} $<br\/> $ =\\dfrac{\\left( {{x}^{2}}+{{y}^{2}} \\right){{\\left( x-y \\right)}^{2}}-{{y}^{2}}{{\\left( x-y \\right)}^{2}}}{\\left( x+y \\right){{x}^{2}}} $<br\/> $ =\\dfrac{{{\\left( x-y \\right)}^{2}}\\left( {{x}^{2}}+{{y}^{2}}-{{y}^{2}} \\right)}{\\left( x+y \\right){{x}^{2}}} $<br\/> $ =\\dfrac{{{\\left( x-y \\right)}^{2}}{{x}^{2}}}{\\left( x+y \\right){{x}^{2}}} $<br\/> $ =\\dfrac{{{\\left( x-y \\right)}^{2}}}{x+y} $ <br\/> Thay $x = y = 10$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{{{\\left( x-y \\right)}^{2}}}{x+y}=\\dfrac{{{\\left( 10-10 \\right)}^{2}}}{10+10}=0$<\/span> "}]}],"id_ques":113},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["15"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/10.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\left( \\dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}-1 \\right)\\cdot \\dfrac{x-y}{2y}$ t\u1ea1i $x=14;y=-15$ l\u00e0 _input_ ","hint":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau \u0111\u00f3 thay $x=14;y=-15$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} &\\left( \\dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}-1 \\right)\\cdot \\dfrac{x-y}{2y}\\\\&=\\left( \\dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}-\\dfrac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}} \\right)\\cdot \\dfrac{x-y}{2y} \\\\ & =\\dfrac{{{x}^{2}}+{{y}^{2}}-{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}\\cdot \\dfrac{x-y}{2y} \\\\ & =\\dfrac{2{{y}^{2}}}{\\left( x+y \\right)\\left( x-y \\right)}\\cdot \\dfrac{x-y}{2y} \\\\ & =\\dfrac{2{{y}^{2}}\\left( x-y \\right)}{\\left( x+y \\right)\\left( x-y \\right)2y} \\\\ & =\\dfrac{y}{x+y} \\\\ \\end{align}$ <br\/> Thay $x=14;y=-15$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{y}{x+y}=\\dfrac{-15}{14-15}=\\dfrac{-15}{-1}=15$ <\/span> "}]}],"id_ques":114},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/9.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\left( {{x}^{2}}-\\dfrac{1}{{{x}^{2}}} \\right):\\left( {{x}^{2}}+\\dfrac{1}{{{x}^{2}}} \\right)$ l\u00e0: ","select":["A. $\\dfrac{x^4+1}{x^4-1}$ ","B. $\\dfrac{x^4-1}{x^4+1}$","C. $x^{16}-1$","D. $x^8-1$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh trong ngo\u1eb7c tr\u01b0\u1edbc. <br\/> <b> B\u01b0\u1edbc 2: <\/b> Th\u1ef1c hi\u1ec7n nh\u00e2n, chia, c\u1ed9ng, tr\u1eeb c\u00e1c ph\u00e2n th\u1ee9c v\u1edbi nhau \u0111\u1ec3 r\u00fat g\u1ecdn. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $\\begin{align} & \\left( {{x}^{2}}-\\dfrac{1}{{{x}^{2}}} \\right):\\left( {{x}^{2}}+\\dfrac{1}{{{x}^{2}}} \\right)\\\\&=\\left( \\dfrac{{{x}^{4}}}{{{x}^{2}}}-\\dfrac{1}{{{x}^{2}}} \\right):\\left( \\dfrac{{{x}^{4}}}{{{x}^{2}}}+\\dfrac{1}{{{x}^{2}}} \\right) \\\\ & =\\dfrac{{{x}^{4}}-1}{{{x}^{2}}}:\\dfrac{{{x}^{4}}+1}{{{x}^{2}}} \\\\ & =\\dfrac{{{x}^{4}}-1}{{{x}^{2}}}\\cdot \\dfrac{{{x}^{2}}}{{{x}^{4}}+1} \\\\ & =\\dfrac{\\left( {{x}^{4}}-1 \\right){{x}^{2}}}{{{x}^{2}}\\left( {{x}^{4}}+1 \\right)} \\\\ & =\\dfrac{{{x}^{4}}-1}{{{x}^{4}}+1} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span> ","column":2}]}],"id_ques":115},{"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/bai14/lv1/img\/8.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{x+y}{z}\\cdot \\left( \\dfrac{x-y}{x}:\\dfrac{{{y}^{2}}}{x+z} \\right)$ l\u00e0: ","select":["A. $\\dfrac{(x^2-y^2)(x+z)}{xy^2z}$ ","B. $\\dfrac{x^2-y^2}{xyz}$","C. $\\dfrac{x+z}{x-y}$","D. $\\dfrac{1}{xyz}$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh trong ngo\u1eb7c tr\u01b0\u1edbc. <br\/> <b> B\u01b0\u1edbc 2: <\/b> Th\u1ef1c hi\u1ec7n nh\u00e2n (chia) c\u00e1c ph\u00e2n th\u1ee9c v\u1edbi nhau \u0111\u1ec3 r\u00fat g\u1ecdn. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{x+y}{z}\\cdot \\left( \\dfrac{x-y}{x}:\\dfrac{{{y}^{2}}}{x+z} \\right) \\\\&=\\dfrac{x+y}{z}\\cdot \\left( \\dfrac{x-y}{x}\\cdot \\dfrac{x+z}{{{y}^{2}}} \\right) \\\\ & =\\dfrac{x+y}{z}\\cdot \\dfrac{\\left( x-y \\right)\\left( x+z \\right)}{x.{{y}^{2}}} \\\\ & =\\dfrac{\\left( x+y \\right)\\left( x-y \\right)\\left( x+z \\right)}{x{{y}^{2}}z} \\\\ & =\\dfrac{\\left( {{x}^{2}}-{{y}^{2}} \\right)\\left( x+z \\right)}{x{{y}^{2}}z} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":2}]}],"id_ques":116},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/5.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\left( \\dfrac{{{x}^{2}}}{y}:\\dfrac{x}{{{y}^{2}}} \\right)\\cdot \\dfrac{{{x}^{2}}}{{{y}^{2}}}$ l\u00e0: ","select":["A. $\\dfrac{x}{y^3}$ ","B. $\\dfrac{{{x}^{3}}}{y}$","C. $\\dfrac{x}{y}$","D. $\\dfrac{{{x}^{2}}}{y^2}$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh trong ngo\u1eb7c tr\u01b0\u1edbc. <br\/> <b> B\u01b0\u1edbc 2: <\/b> Th\u1ef1c hi\u1ec7n nh\u00e2n (chia) c\u00e1c ph\u00e2n th\u1ee9c v\u1edbi nhau \u0111\u1ec3 r\u00fat g\u1ecdn. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $\\begin{align} &\\left( \\dfrac{{{x}^{2}}}{y}:\\dfrac{x}{{{y}^{2}}} \\right)\\cdot \\dfrac{{{x}^{2}}}{{{y}^{2}}}\\\\&=\\left( \\dfrac{{{x}^{2}}}{y}\\cdot \\dfrac{{{y}^{2}}}{x} \\right)\\cdot \\dfrac{{{x}^{2}}}{{{y}^{2}}} \\\\ & =\\left( \\dfrac{{{x}^{2}}.{{y}^{2}}}{y.x} \\right)\\cdot \\dfrac{{{x}^{2}}}{{{y}^{2}}} \\\\ & =\\dfrac{{{x}^{2}}{{y}^{2}}}{xy}\\cdot \\dfrac{{{x}^{2}}}{{{y}^{2}}} \\\\ & =\\dfrac{xy}{1}\\cdot \\dfrac{{{x}^{2}}}{{{y}^{2}}} \\\\ & =\\dfrac{xy.{{x}^{2}}}{1.{{y}^{2}}} \\\\ & =\\dfrac{{{x}^{3}}y}{{{y}^{2}}} \\\\ & =\\dfrac{{{x}^{3}}}{y} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span> ","column":2}]}],"id_ques":117},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p nh\u00e2n (chia) c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank_random","correct":[[["6xy"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/4.jpg' \/><\/center> $\\dfrac{6{{x}^{3}}}{{{y}^{2}}}:\\dfrac{{{x}^{2}}}{{{y}^{3}}}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> Chia hai ph\u00e2n th\u1ee9c:<\/b> <br\/> $\\dfrac{A}{B} : \\dfrac{C}{D}=\\dfrac{A}{B} \\cdot \\dfrac{D}{C}=\\dfrac{A.D}{B.C}$ v\u1edbi $B, D, \\dfrac{C}{D} \\ne 0$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $\\dfrac{6{{x}^{3}}}{{{y}^{2}}}:\\dfrac{{{x}^{2}}}{{{y}^{3}}}$$=\\dfrac{6{{x}^{3}}}{{{y}^{2}}}\\cdot \\dfrac{{{y}^{3}}}{{{x}^{2}}}$$=\\dfrac{6{{x}^{3}}.{{y}^{3}}}{{{y}^{2}}.{{x}^{2}}}$$=\\dfrac{6{{x}^{3}}{{y}^{3}}}{{{x}^{2}}{{y}^{2}}}=6xy$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $6xy$ <\/span><\/span> "}]}],"id_ques":118},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p nh\u00e2n (chia) c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["x+1"],["x-1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/3.jpg' \/><\/center> $\\dfrac{{{x}^{2}}+2x+1}{{{x}^{3}}-1}\\cdot \\dfrac{2{{x}^{2}}+2x+2}{x+1}$$=\\dfrac{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]{}}$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> Nh\u00e2n hai ph\u00e2n th\u1ee9c:<\/b> <br\/> $\\dfrac{A}{B} \\cdot \\dfrac{C}{D}=\\dfrac{A.C}{B.D}$ v\u1edbi $B, D \\ne 0$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $\\begin{align} &\\dfrac{{{x}^{2}}+2x+1}{{{x}^{3}}-1}\\cdot \\dfrac{2{{x}^{2}}+2x+2}{x+1}\\\\&=\\dfrac{\\left( {{x}^{2}}+2x+1 \\right).\\left( 2{{x}^{2}}+2x+2 \\right)}{\\left( {{x}^{3}}-1 \\right)\\left( x+1 \\right)} \\\\ & =\\dfrac{{{\\left( x+1 \\right)}^{2}}.2\\left( {{x}^{2}}+x+1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)\\left( x+1 \\right)} \\\\ & =\\dfrac{2\\left( x+1 \\right)}{x-1} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{2\\left( x+1 \\right)}{x-1}$ <\/span><\/span> "}]}],"id_ques":119},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p nh\u00e2n (chia) c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["30"],["7xy"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv1/img\/2.jpg' \/><\/center> $\\dfrac{15x}{7{{y}^{3}}}\\cdot \\dfrac{2{{y}^{2}}}{{{x}^{2}}}=\\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]{}}$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> Nh\u00e2n hai ph\u00e2n th\u1ee9c:<\/b> <br\/> $\\dfrac{A}{B} \\cdot \\dfrac{C}{D}=\\dfrac{A.C}{B.D}$ v\u1edbi $B, D \\ne 0$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $\\dfrac{15x}{7{{y}^{3}}}\\cdot \\dfrac{2{{y}^{2}}}{{{x}^{2}}}$$=\\dfrac{15x.2{{y}^{2}}}{7{{y}^{3}}.{{x}^{2}}}$$=\\dfrac{30x{{y}^{2}}}{7{{x}^{2}}{{y}^{3}}}=\\dfrac{30}{7xy}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{30}{7xy}$ <\/span><\/span> "}]}],"id_ques":120}],"lesson":{"save":0,"level":1}}