{"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":[[["8xy"],["1"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/16.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ M=\\dfrac{64{{x}^{2}}{{y}^{2}}-1}{{{x}^{2}}-4}$$\\cdot \\dfrac{{{\\left( x+2 \\right)}^{2}}}{{{x}^{2}}-4}$$\\cdot \\dfrac{{{\\left( x-2 \\right)}^{2}}}{8xy+1}$ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $M$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $M = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} - \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & M=\\dfrac{64{{x}^{2}}{{y}^{2}}-1}{{{x}^{2}}-4}\\cdot \\dfrac{{{\\left( x+2 \\right)}^{2}}}{{{x}^{2}}-4}\\cdot \\dfrac{{{\\left( x-2 \\right)}^{2}}}{8xy+1} \\\\ & =\\dfrac{\\left( 8xy+1 \\right)\\left( 8xy-1 \\right)}{\\left( x+2 \\right)\\left( x-2 \\right)}\\cdot \\dfrac{{{\\left( x+2 \\right)}^{2}}}{\\left( x+2 \\right)\\left( x-2 \\right)}\\cdot \\dfrac{{{\\left( x-2 \\right)}^{2}}}{8xy+1} \\\\ & =\\dfrac{\\left( 8xy+1 \\right)\\left( 8xy-1 \\right){{\\left( x+2 \\right)}^{2}}{{\\left( x-2 \\right)}^{2}}}{{{\\left( x+2 \\right)}^{2}}{{\\left( x-2 \\right)}^{2}}\\left( 8xy+1 \\right)} \\\\ & =\\dfrac{8xy-1}{1} \\\\ & =8xy-1 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $8xy$ v\u00e0 $1$. <\/span><\/span> "}]}],"id_ques":141},{"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":[[["511"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/16.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ M=\\dfrac{64{{x}^{2}}{{y}^{2}}-1}{{{x}^{2}}-4}$$\\cdot \\dfrac{{{\\left( x+2 \\right)}^{2}}}{{{x}^{2}}-4}$$\\cdot \\dfrac{{{\\left( x-2 \\right)}^{2}}}{8xy+1}$ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> V\u1edbi $x=y=8$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $M$ l\u00e0 _input_ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1, ta c\u00f3: $M=8xy-1$ <br\/> Thay $x=y=8$ v\u00e0o bi\u1ec3u th\u1ee9c $M$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $M=8xy-1=8.8.8-1=511$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $511$. <\/span><\/span> "}]}],"id_ques":142},{"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":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/11.jpg' \/><\/center> $\\dfrac{{{x}^{7}}+{{x}^{5}}+1}{{{x}^{3}}-1}$$\\cdot \\dfrac{x-1}{x+2}$$\\cdot \\dfrac{{{x}^{2}}+x+1}{{{x}^{7}}+{{x}^{5}}+1}$$=\\dfrac{1}{x+2}$ ","select":["\u0110\u00fang","Sai"],"hint":" \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t giao ho\u00e1n c\u1ee7a ph\u00e9p nh\u00e2n \u0111\u1ec3 nh\u00f3m c\u00e1c th\u1eeba s\u1ed1 c\u00f3 th\u1ec3 r\u00fat g\u1ecdn \u0111\u01b0\u1ee3c","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> V\u1ebf tr\u00e1i = $ \\dfrac{{{x}^{7}}+{{x}^{5}}+1}{{{x}^{3}}-1}\\cdot \\dfrac{x-1}{x+2}$$\\cdot \\dfrac{{{x}^{2}}+x+1}{{{x}^{7}}+{{x}^{5}}+1} $<br\/>$ =\\left( \\dfrac{{{x}^{7}}+{{x}^{5}}+1}{{{x}^{3}}-1}\\cdot \\dfrac{{{x}^{2}}+x+1}{{{x}^{7}}+{{x}^{5}}+1} \\right)$$\\cdot \\dfrac{x-1}{x+2} $<br\/>$ =\\dfrac{{{x}^{2}}+x+1}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)}$$\\cdot \\dfrac{x-1}{x+2} $<br\/>$ =\\dfrac{1}{x-1}\\cdot \\dfrac{x-1}{x+2} $<br\/>$ =\\dfrac{1}{x+2}$ = V\u1ebf ph\u1ea3i. <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":143},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/10.jpg' \/><\/center> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{\\dfrac{2{{x}^{2}}+3}{x}}{\\,\\,\\,\\,\\,x+1\\,\\,\\,\\,\\,\\,\\,}$ l\u00e0:","select":["A. $x\\in \\mathbb{R}$ ","B. $x\\ne \\{-1;0\\}$ ","C. $x\\ne 0$","D. $x\\ne \\{-1;0;1\\}$"],"explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{\\dfrac{2{{x}^{2}}+3}{x}}{\\,\\,\\,\\,\\,x+1\\,\\,\\,\\,\\,\\,\\,}$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & x+1\\ne 0 \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & x\\ne -1 \\\\ \\end{aligned} \\right.$ <br\/> V\u1eady bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a khi $x\\ne \\{-1;0\\}$ <br\/> <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":144},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/9.jpg' \/><\/center> Cho bi\u1ec3u th\u1ee9c $\\dfrac{a}{b-2}-\\dfrac{{{a}^{2}}+2a+1}{{{b}^{2}}-4}\\cdot P$$=-\\dfrac{1}{b-2}.$ Khi \u0111\u00f3 ph\u00e2n th\u1ee9c $P$ l\u00e0: ","select":["A. $\\dfrac{a+2}{b+1}$ ","B. $\\dfrac{b-2}{a+1}$ ","C. $\\dfrac{1}{a+1}$","D. $\\dfrac{b+2}{a+1}$"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{a}{b-2}-\\dfrac{{{a}^{2}}+2a+1}{{{b}^{2}}-4}\\cdot P=-\\dfrac{1}{b-2} \\\\ &\\Leftrightarrow \\dfrac{a}{b-2}-\\dfrac{{{\\left( a+1 \\right)}^{2}}}{\\left( b+2 \\right)\\left( b-2 \\right)}\\cdot P=-\\dfrac{1}{b-2} \\\\ &\\Leftrightarrow \\dfrac{{{\\left( a+1 \\right)}^{2}}}{\\left( b+2 \\right)\\left( b-2 \\right)}\\cdot P=\\dfrac{a}{b-2}+\\dfrac{1}{b-2} \\\\ &\\Leftrightarrow \\dfrac{{{\\left( a+1 \\right)}^{2}}}{\\left( b+2 \\right)\\left( b-2 \\right)}\\cdot P=\\dfrac{a+1}{b-2} \\\\ & \\Leftrightarrow P=\\dfrac{a+1}{b-2}:\\dfrac{{{\\left( a+1 \\right)}^{2}}}{\\left( b+2 \\right)\\left( b-2 \\right)} \\\\ & =\\dfrac{a+1}{b-2}\\cdot \\dfrac{\\left( b+2 \\right)\\left( b-2 \\right)}{{{\\left( a+1 \\right)}^{2}}} \\\\ & =\\dfrac{\\left( a+1 \\right)\\left( b+2 \\right)\\left( b-2 \\right)}{\\left( b-2 \\right){{\\left( a+1 \\right)}^{2}}} \\\\ & =\\dfrac{b+2}{a+1} \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":145},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/8.jpg' \/><\/center> Cho bi\u1ec3u th\u1ee9c $\\left( 2-P \\right)$$:\\left\\{ \\dfrac{{{m}^{2}}-{{a}^{2}}}{{{m}^{3}}+{{a}^{3}}}\\cdot \\left[ \\left( m-\\dfrac{{{m}^{2}}+{{a}^{2}}}{a} \\right):\\left( \\dfrac{1}{m}-\\dfrac{1}{a} \\right) \\right] \\right\\}$$=1.$ Khi \u0111\u00f3 ph\u00e2n th\u1ee9c $P$ l\u00e0:","select":["A. $m+2$ ","B. $2-m$ ","C. $a+m$","D. $2-a$"],"hint":"Ta r\u00fat g\u1ecdn c\u00e1c bi\u1ec3u th\u1ee9c trong ngo\u1eb7c () tr\u01b0\u1edbc, sau \u0111\u00f3 ngo\u1eb7c [], cu\u1ed1i c\u00f9ng l\u00e0 ngo\u1eb7c {} ","explain":"<span class='basic_left'>Ta c\u00f3: $ \\left( 2-P \\right)$$:\\left\\{ \\dfrac{{{m}^{2}}-{{a}^{2}}}{{{m}^{3}}+{{a}^{3}}}\\cdot \\left[ \\left( m-\\dfrac{{{m}^{2}}+{{a}^{2}}}{a} \\right):\\left( \\dfrac{1}{m}-\\dfrac{1}{a} \\right) \\right] \\right\\}$$=1 $ <br\/> $\\Leftrightarrow \\left( 2-P \\right)$$:\\left\\{ \\dfrac{{{m}^{2}}-{{a}^{2}}}{{{m}^{3}}+{{a}^{3}}}\\cdot \\left[ \\left( \\dfrac{ma-{{m}^{2}}-{{a}^{2}}}{a} \\right):\\left( \\dfrac{a-m}{ma} \\right) \\right] \\right\\}$$=1 $ <br\/> $\\Leftrightarrow \\left( 2-P \\right)$$:\\left\\{ \\dfrac{{{m}^{2}}-{{a}^{2}}}{{{m}^{3}}+{{a}^{3}}}\\cdot \\left[ \\dfrac{-\\left( {{m}^{2}}-ma+{{a}^{2}} \\right)}{a}\\cdot \\dfrac{ma}{a-m} \\right] \\right\\}$$=1 $ <br\/> $\\Leftrightarrow \\left( 2-P \\right)$$:\\left\\{ \\dfrac{\\left( m+a \\right)\\left( m-a \\right)}{\\left( m+a \\right)\\left( {{m}^{2}}-ma+{{a}^{2}} \\right)}\\cdot \\left[ \\dfrac{-m\\left( {{m}^{2}}-ma+{{a}^{2}} \\right)}{a-m} \\right] \\right\\}$$=1 $<br\/>$\\Leftrightarrow \\left( 2-P \\right)$$:\\left[ -\\dfrac{\\left( m-a \\right)m\\left( {{m}^{2}}-ma+{{a}^{2}} \\right)}{\\left( {{m}^{2}}-ma+{{a}^{2}} \\right)\\left( a-m \\right)} \\right]$$=1 $ <br\/> $\\Leftrightarrow \\left( 2-P \\right)$$:\\left[ \\dfrac{\\left( a-m \\right)m}{a-m} \\right]$$=1 $ <br\/> $\\Leftrightarrow \\left( 2-P \\right):m=1 $<br\/>$\\Leftrightarrow 2-P=m $ <br\/> $\\Leftrightarrow P=2-m$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":146},{"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":10,"width":50,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/5.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{{{x}^{4}}-x{{y}^{3}}}{2xy+{{y}^{2}}}:\\dfrac{{{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}}{2x+y}$ t\u1ea1i $x = 2;y=1$ l\u00e0 _input_ ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{{{x}^{4}}-x{{y}^{3}}}{2xy+{{y}^{2}}}:\\dfrac{{{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}}{2x+y} \\\\ & =\\dfrac{{{x}^{4}}-x{{y}^{3}}}{2xy+{{y}^{2}}}\\cdot \\dfrac{2x+y}{{{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}} \\\\ & =\\dfrac{\\left( {{x}^{4}}-x{{y}^{3}} \\right)\\left( 2x+y \\right)}{\\left( 2xy+{{y}^{2}} \\right)\\left( {{x}^{3}}+{{x}^{2}}y+x{{y}^{2}} \\right)} \\\\ & =\\dfrac{x\\left( {{x}^{3}}-{{y}^{3}} \\right)\\left( 2x+y \\right)}{y\\left( 2x+y \\right)x\\left( {{x}^{2}}+xy+{{y}^{2}} \\right)} \\\\ & =\\dfrac{x\\left( x-y \\right)\\left( {{x}^{2}}+xy+{{y}^{2}} \\right)\\left( 2x+y \\right)}{y\\left( 2x+y \\right)x\\left( {{x}^{2}}+xy+{{y}^{2}} \\right)} \\\\ & =\\dfrac{x-y}{y} \\\\ \\end{align}$ <br\/> Thay $x = 2;y=1$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{x-y}{y}=\\dfrac{2-1}{1}=1$<\/span>"}]}],"id_ques":147},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/4.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{{{x}^{2}}-4}{x+10}\\cdot \\dfrac{x}{x+2}$$+\\dfrac{{{x}^{2}}-4}{x+10}\\cdot \\dfrac{1-x}{x+2}$ l\u00e0: ","select":["A. $\\dfrac{x}{\\left( x+10 \\right)\\left( x+2 \\right)}$ ","B. $\\dfrac{x-2}{ x+10}$","C. $\\dfrac{x-2}{\\left( x+10 \\right)\\left( x+2 \\right)}$","D. $\\dfrac{x+2}{x+10}$"],"explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $ \\dfrac{{{x}^{2}}-4}{x+10}\\cdot \\dfrac{x}{x+2}$$+\\dfrac{{{x}^{2}}-4}{x+10}\\cdot \\dfrac{1-x}{x+2} $<br\/>$ =\\dfrac{\\left( {{x}^{2}}-4 \\right)x}{\\left( x+10 \\right)\\left( x+2 \\right)}$$+\\dfrac{\\left( {{x}^{2}}-4 \\right)\\left( 1-x \\right)}{\\left( x+10 \\right)\\left( x+2 \\right)} $<br\/>$ =\\dfrac{\\left( {{x}^{2}}-4 \\right)x+\\left( {{x}^{2}}-4 \\right)\\left( 1-x \\right)}{\\left( x+10 \\right)\\left( x+2 \\right)} $<br\/>$ =\\dfrac{(x^2-4)(x+1-x)}{\\left( x+10 \\right)\\left( x+2 \\right)} $<br\/>$ =\\dfrac{{{x}^{2}}-4}{\\left( x+10 \\right)\\left( x+2 \\right)} $ <br\/>$ =\\dfrac{(x-2)(x+2)}{\\left( x+10 \\right)\\left( x+2 \\right)} $ <br\/>$ =\\dfrac{x-2}{ x+10 } $ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span> ","column":2}]}],"id_ques":148},{"time":24,"part":[{"title":"\u0110i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng trong k\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p nh\u00e2n (chia) c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["b"],["a"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/3.jpg' \/><\/center> $\\dfrac{{{\\left( a+b \\right)}^{2}}}{ab-{{b}^{2}}}:\\left[ -\\dfrac{ab+{{b}^{2}}}{{{\\left( a-b \\right)}^{2}}} \\right]$$=\\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]{}}^{2}}}{{{b}^{2}}}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{{{\\left( a+b \\right)}^{2}}}{ab-{{b}^{2}}}:\\left[ -\\dfrac{ab+{{b}^{2}}}{{{\\left( a-b \\right)}^{2}}} \\right] \\\\ & =\\dfrac{{{\\left( a+b \\right)}^{2}}}{ab-{{b}^{2}}}\\cdot \\left[ -\\dfrac{{{\\left( a-b \\right)}^{2}}}{ab+{{b}^{2}}} \\right] \\\\ & =\\dfrac{{{\\left( a+b \\right)}^{2}}}{b\\left( a-b \\right)}\\cdot \\left[ -\\dfrac{{{\\left( a-b \\right)}^{2}}}{b\\left( a+b \\right)} \\right] \\\\ & =-\\dfrac{{{\\left( a+b \\right)}^{2}}{{\\left( a-b \\right)}^{2}}}{{{b}^{2}}\\left( a-b \\right)\\left( a+b \\right)} \\\\ & =-\\dfrac{\\left( a+b \\right)\\left( a-b \\right)}{{{b}^{2}}} \\\\ & =\\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}}$ <\/span><\/span> "}]}],"id_ques":149},{"time":24,"part":[{"title":"\u0110i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng trong k\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p nh\u00e2n (chia) c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["6a"],["x-y"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv3/img\/2.jpg' \/><\/center> $\\dfrac{a\\,{{x}^{2}}-a{{y}^{2}}}{{{x}^{2}}+2xy+{{y}^{2}}}\\cdot \\dfrac{6{{x}^{3}}+6{{y}^{3}}}{{{x}^{2}}-2xy+{{y}^{2}}}$$=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\,\\left( {{x}^{2}}-xy+{{y}^{2}} \\right)}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} }$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{a\\,{{x}^{2}}-a{{y}^{2}}}{{{x}^{2}}+2xy+{{y}^{2}}}\\cdot \\dfrac{6{{x}^{3}}+6{{y}^{3}}}{{{x}^{2}}-2xy+{{y}^{2}}} \\\\ & =\\dfrac{a\\left( {{x}^{2}}-{{y}^{2}} \\right)}{{{\\left( x+y \\right)}^{2}}}\\cdot \\dfrac{6\\left( {{x}^{3}}+{{y}^{3}} \\right)}{{{\\left( x-y \\right)}^{2}}} \\\\ & =\\dfrac{a\\left( {{x}^{2}}-{{y}^{2}} \\right)6\\left( {{x}^{3}}+{{y}^{3}} \\right)}{{{\\left( x+y \\right)}^{2}}{{\\left( x-y \\right)}^{2}}} \\\\ & =\\dfrac{6a\\left( x+y \\right)\\left( x-y \\right)\\left( x+y \\right)\\left( {{x}^{2}}-xy+{{y}^{2}} \\right)}{{{\\left( x+y \\right)}^{2}}{{\\left( x-y \\right)}^{2}}} \\\\ & =\\dfrac{6a{{\\left( x+y \\right)}^{2}}\\left( x-y \\right)\\left( {{x}^{2}}-xy+{{y}^{2}} \\right)}{{{\\left( x+y \\right)}^{2}}{{\\left( x-y \\right)}^{2}}} \\\\ & =\\dfrac{6a\\left( {{x}^{2}}-xy+{{y}^{2}} \\right)}{\\left( x-y \\right)} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{6a\\left( {{x}^{2}}-xy+{{y}^{2}} \\right)}{\\left( x-y \\right)}$ <\/span><\/span> "}]}],"id_ques":150}],"lesson":{"save":0,"level":3}}