{"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/lv2/img\/11.jpg' \/><\/center> <span class='basic_left'> Cho bi\u1ec3u th\u1ee9c $ A=\\dfrac{\\left( x+1 \\right)\\left( {{x}^{2}}-2x+1 \\right)}{6{{x}^{3}}+6}$ $:$ $\\dfrac{{{x}^{2}}-1}{4{{x}^{2}}-4x+4}$ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $A$ l\u00e0: $\\left\\{ \\begin{align} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $A$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & 6{{x}^{3}}+6\\ne 0 \\\\ & 4{{x}^{2}}-4x+4\\ne 0 \\\\ & {{x}^{2}}-1\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow\\left\\{ \\begin{aligned} & {{x}^{3}}+1\\ne 0 \\\\ & {{x}^{2}}-x+1\\ne 0 \\,\\, \\forall x \\\\ & {{x}^{2}}\\ne 1 \\\\ \\end{aligned} \\right. \\Rightarrow \\left\\{ \\begin{aligned} & x\\ne -1 \\\\ & x\\ne 1 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ v\u00e0 $-1$. <\/span><\/span> "}]}],"id_ques":121},{"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":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv2/img\/11.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ A=\\dfrac{\\left( x+1 \\right)\\left( {{x}^{2}}-2x+1 \\right)}{6{{x}^{3}}+6}$ $:$ $\\dfrac{{{x}^{2}}-1}{4{{x}^{2}}-4x+4}$ <br\/> <br\/> <br\/> <b> C\u00e2u 2 : <\/b> \u0110\u1ec3 $A=0$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0: <\/span> ","select":["A. $x = 1$ ","B. $x = \\pm 1$ ","C. Kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o th\u1ecfa m\u00e3n "],"hint":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A.$ <br\/> Cho $A=0$ \u0111\u1ec3 t\u00ecm $x.$ ","explain":"<span class='basic_left'> Theo c\u00e2u 1, \u0111i\u1ec1u ki\u1ec7n: $ \\left\\{ \\begin{align} & x\\ne -1 \\\\ & x\\ne 1 \\\\ \\end{align} \\right.$ <br\/> Ta c\u00f3:<br\/> $ A=\\dfrac{\\left( x+1 \\right)\\left( {{x}^{2}}-2x+1 \\right)}{6{{x}^{3}}+6}$ $: $ $\\dfrac{{{x}^{2}}-1}{4{{x}^{2}}-4x+4}$ <br\/> $ =\\dfrac{\\left( x+1 \\right){{\\left( x-1 \\right)}^{2}}}{6\\left( {{x}^{3}}+1 \\right)} $ $: $ $\\dfrac{\\left( x+1 \\right)\\left( x-1 \\right)}{4\\left( {{x}^{2}}-x+1 \\right)} $ <br\/> $ =\\dfrac{\\left( x+1 \\right){{\\left( x-1 \\right)}^{2}}}{6\\left( x+1 \\right)\\left( {{x}^{2}}-x+1 \\right)} $ $\\cdot \\dfrac{4\\left( {{x}^{2}}-x+1 \\right)}{\\left( x+1 \\right)\\left( x-1 \\right)} $ <br\/> $ =\\dfrac{2\\left( x-1 \\right)}{3\\left( x+1 \\right)} $<br\/> \u0110\u1ec3 $ A = 0 \\Leftrightarrow \\dfrac{2\\left( x-1 \\right)}{3\\left( x+1 \\right)}=0 $ <br\/> $\\Leftrightarrow 2(x-1)=0$<br\/> $\\Leftrightarrow x=1$ (kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n) <br\/> V\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $x$ \u0111\u1ec3 $A = 0.$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span>","column":1}]}],"id_ques":122},{"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/lv2/img\/10.jpg' \/><\/center> $ \\dfrac{2x+y}{3x+y}\\cdot \\dfrac{4x-3y}{x-y}$$-$$\\dfrac{2x+y}{3x+y}\\cdot \\dfrac{x-4y}{x-y} $$=\\dfrac{2xy}{x-y} $ ","select":["\u0110\u00fang","Sai"],"hint":" Ta th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh v\u1ebf tr\u00e1i r\u1ed3i so s\u00e1nh k\u1ebft qu\u1ea3 v\u1edbi v\u1ebf ph\u1ea3i. ","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> V\u1ebf tr\u00e1i = $ \\dfrac{2x+y}{3x+y}\\cdot \\dfrac{4x-3y}{x-y}$$-\\dfrac{2x+y}{3x+y}\\cdot \\dfrac{x-4y}{x-y} $ <br\/>$=\\dfrac{\\left( 2x+y \\right)\\left( 4x-3y \\right)}{\\left( 3x+y \\right)\\left( x-y \\right)}$$-\\dfrac{\\left( 2x+y \\right)\\left( x-4y \\right)}{\\left( 3x+y \\right)\\left( x-y \\right)}$<br\/> $ =\\dfrac{\\left( 2x+y \\right)\\left( 4x-3y \\right)-\\left( 2x+y \\right)\\left( x-4y \\right)}{\\left( 3x+y \\right)\\left( x-y \\right)} $<br\/>$ =\\dfrac{\\left( 2x+y \\right)\\left( 4x-3y-x+4y \\right)}{\\left( 3x+y \\right)\\left( x-y \\right)}$ <br\/> $ =\\dfrac{\\left( 2x+y \\right)\\left( 3x+y \\right)}{\\left( 3x+y \\right)\\left( x-y \\right)} $ <br\/>$ =\\dfrac{2x+y}{x-y} \\ne$ V\u1ebf ph\u1ea3i <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":123},{"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/lv2/img\/9.jpg' \/><\/center> $\\dfrac{8{{\\left( x-y \\right)}^{2}}{{\\left( x+z \\right)}^{2}}}{5{{\\left( x+z \\right)}^{4}}}$ $\\cdot \\dfrac{{{\\left( x+z \\right)}^{3}}}{16{{\\left( x-y \\right)}^{4}}} $ $=\\dfrac{x+z}{10{{\\left( x-y \\right)}^{2}}}$ ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> V\u1ebf tr\u00e1i = $ \\dfrac{8{{\\left( x-y \\right)}^{2}}{{\\left( x+z \\right)}^{2}}}{5{{\\left( x+z \\right)}^{4}}} $ $\\cdot \\dfrac{{{\\left( x+z \\right)}^{3}}}{16{{\\left( x-y \\right)}^{4}}} $<br\/> $ =\\dfrac{8{{\\left( x-y \\right)}^{2}}{{\\left( x+z \\right)}^{2}}{{\\left( x+z \\right)}^{3}}}{5{{\\left( x+z \\right)}^{4}}.16{{\\left( x-y \\right)}^{4}}} $<br\/> $ =\\dfrac{x+z}{10{{\\left( x-y \\right)}^{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":124},{"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/lv2/img\/8.jpg' \/><\/center> $\\left( \\dfrac{b}{{{a}^{2}}-ab}-\\dfrac{a}{ab-{{b}^{2}}} \\right)\\cdot \\dfrac{{{a}^{2}}b-a{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}=-1$ ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> V\u1ebf tr\u00e1i = $ \\left( \\dfrac{b}{{{a}^{2}}-ab}-\\dfrac{a}{ab-{{b}^{2}}} \\right)\\cdot \\dfrac{{{a}^{2}}b-a{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}} $ <br\/> $=\\left( \\dfrac{b}{a\\left( a-b \\right)}-\\dfrac{a}{b\\left( a-b \\right)} \\right)\\cdot \\dfrac{{{a}^{2}}b-a{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}} $ <br\/> $ =\\left( \\dfrac{{{b}^{2}}}{ab\\left( a-b \\right)}-\\dfrac{{{a}^{2}}}{ab\\left( a-b \\right)} \\right)\\cdot \\dfrac{{{a}^{2}}b-a{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}} $ <br\/> $ =\\dfrac{{{b}^{2}}-{{a}^{2}}}{ab\\left( a-b \\right)}\\cdot \\dfrac{ab\\left( a-b \\right)}{{{a}^{2}}-{{b}^{2}}} $ <br\/> $ =-1 =$ V\u1ebf ph\u1ea3i <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":125},{"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/lv2/img\/5.jpg' \/><\/center> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{x-1}{x-3}:\\dfrac{x-3}{x-4}:\\dfrac{x-4}{x-5}$ l\u00e0:","select":["A. $x\\ne \\{3;4;5\\}$ ","B. $x\\ne \\{1;3;4;5\\}$ ","C. $x\\ne \\{3;4\\}$","D. $x > 5$"],"explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{x-1}{x-3}:\\dfrac{x-3}{x-4}:\\dfrac{x-4}{x-5}$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x-3\\ne 0 \\\\ & x-4\\ne 0 \\\\ & x-5\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 3 \\\\ & x\\ne 4 \\\\ & x\\ne 5 \\\\ \\end{aligned} \\right.$ <br\/> <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 l\u00e0 $B;D\\,\\,\\ne 0;\\,\\,\\dfrac{C}{D}\\,\\,\\ne 0$ .<br\/> Do \u0111\u00f3 ta \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $\\dfrac{A}{B}:\\dfrac{C}{D}$ c\u00f3 ngh\u0129a, ta vi\u1ebft g\u1ecdn l\u1ea1i l\u00e0: $\\left\\{ \\begin{align} & B\\ne 0 \\\\ & D\\ne 0 \\\\ & C\\ne 0 \\\\ \\end{align} \\right.$","column":2}]}],"id_ques":126},{"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/lv2/img\/4.jpg' \/><\/center> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{2-x}{x+1}\\cdot \\left( 1-\\dfrac{1}{1-x} \\right)$ l\u00e0:","select":["A. $x\\ne \\{-1;0;1\\}$ ","B. $x\\ne \\{-1;1\\}$ ","C. $x\\ne \\{0;1\\}$","D. $x\\ne \\{-1;1;2\\}$"],"explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{2-x}{x+1}\\cdot \\left( 1-\\dfrac{1}{1-x} \\right)$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x+1\\ne 0 \\\\ & 1-x\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne -1 \\\\ & x\\ne 1 \\\\ \\end{aligned} \\right.$ <br\/> <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":127},{"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/lv2/img\/3.jpg' \/><\/center> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{\\left( x+1 \\right)\\left( {{x}^{2}}-2x+1 \\right)}{6{{x}^{3}}+6}$ $:\\dfrac{{{x}^{2}}-1}{4{{x}^{2}}-4x+1}$ l\u00e0:","select":["A. $x\\ne \\{-1;1\\}$ ","B. $x\\ne \\left\\{-1;\\dfrac{1}{2};1\\right\\}$ ","C. $x\\ne \\left\\{-1;\\dfrac{1}{2}\\right\\}$","D. $x\\ne \\left\\{-1;1;-\\dfrac{1}{2};\\dfrac{1}{2}\\right\\}$"],"explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{\\left( x+1 \\right)\\left( {{x}^{2}}-2x+1 \\right)}{6{{x}^{3}}+6}:\\dfrac{{{x}^{2}}-1}{4{{x}^{2}}-4x+1}$ l\u00e0:<br\/> $ \\left\\{ \\begin{align} & 6{{x}^{3}}+6\\ne 0 \\\\ & 4{{x}^{2}}-4x+1\\ne 0 \\\\ & {{x}^{2}}-1\\ne 0 \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\left\\{ \\begin{align} & 6\\left( {{x}^{3}}+1 \\right)\\ne 0 \\\\ & {{\\left( 2x-1 \\right)}^{2}}\\ne 0 \\\\ & \\left( x+1 \\right)\\left( x-1 \\right)\\ne 0 \\\\ \\end{align} \\right. $ <br\/> $ \\Rightarrow \\left\\{ \\begin{align} & x\\ne -1 \\\\ & x\\ne \\dfrac{1}{2} \\\\ & x\\ne 1 \\\\ \\end{align} \\right. $ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span> <br\/> <b> L\u01b0u \u00fd:<\/b> Trong ph\u00e9p chia $\\dfrac{A}{B}:\\dfrac{C}{D}$ th\u00ec \u0111i\u1ec1u ki\u1ec7n l\u00e0 $B;D\\,\\,\\ne 0;\\,\\,\\dfrac{C}{D}\\,\\,\\ne 0$ .<br\/> Do \u0111\u00f3 ta \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $\\dfrac{A}{B}:\\dfrac{C}{D}$ c\u00f3 ngh\u0129a, ta vi\u1ebft g\u1ecdn l\u1ea1i l\u00e0: $\\left\\{ \\begin{align} & B\\ne 0 \\\\ & D\\ne 0 \\\\ & C\\ne 0 \\\\ \\end{align} \\right.$ <br\/> <span><br\/> ","column":2}]}],"id_ques":128},{"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/lv2/img\/2.jpg' \/><\/center> Cho bi\u1ec3u th\u1ee9c $\\dfrac{a+b}{a-b}\\cdot P=\\dfrac{{{a}^{2}}+ab}{2{{a}^{2}}-2{{b}^{2}}}.$ Khi \u0111\u00f3 ph\u00e2n th\u1ee9c $P$ l\u00e0:","select":["A. $\\dfrac{a}{2}$ ","B. $\\dfrac{b}{a-b}$ ","C. $\\dfrac{a}{2(a+b)}$","D. $\\dfrac{a}{a-b}$"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{a+b}{a-b}\\cdot P=\\dfrac{{{a}^{2}}+ab}{2{{a}^{2}}-2{{b}^{2}}} \\\\ & \\Rightarrow P=\\dfrac{{{a}^{2}}+ab}{2{{a}^{2}}-2{{b}^{2}}}:\\dfrac{a+b}{a-b} \\\\ & =\\dfrac{a\\left( a+b \\right)}{2\\left( a+b \\right)\\left( a-b \\right)}\\cdot \\dfrac{a-b}{a+b} \\\\ & =\\dfrac{a}{2\\left( a+b \\right)} \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span>","column":2}]}],"id_ques":129},{"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/lv2/img\/16.jpg' \/><\/center> Cho bi\u1ec3u th\u1ee9c $\\dfrac{a+1}{{{a}^{3}}-1}\\cdot P=\\dfrac{2a+2}{{{a}^{2}}+a+1}.$ Khi \u0111\u00f3 ph\u00e2n th\u1ee9c $P$ l\u00e0:","select":["A. $2(a-1)$ ","B. $a-1$ ","C. $a+1$","D. $\\dfrac{1}{a-1}$"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{a+1}{{{a}^{3}}-1}\\cdot P=\\dfrac{2a+2}{{{a}^{2}}+a+1} \\\\ &\\Rightarrow P=\\dfrac{2a+2}{{{a}^{2}}+a+1}:\\dfrac{a+1}{{{a}^{3}}-1} \\\\ & =\\dfrac{2\\left( a+1 \\right)}{{{a}^{2}}+a+1}:\\dfrac{a+1}{\\left( a-1 \\right)\\left( {{a}^{2}}+a+1 \\right)} \\\\ & =\\dfrac{2\\left( a+1 \\right)}{{{a}^{2}}+a+1}\\cdot \\dfrac{\\left( a-1 \\right)\\left( {{a}^{2}}+a+1 \\right)}{a+1} \\\\ & =2\\left( a-1 \\right) \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":130},{"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/lv2/img\/13.jpg' \/><\/center> Cho bi\u1ec3u th\u1ee9c $\\dfrac{{{a}^{2}}-2ab}{{{a}^{2}}b}\\cdot P=\\dfrac{{{a}^{2}}b-4{{b}^{3}}}{3a{{b}^{2}}}.$ Khi \u0111\u00f3 ph\u00e2n th\u1ee9c $P$ l\u00e0:","select":["A. $\\dfrac{a+b}{3}$ ","B. $\\dfrac{3}{a+b}$ ","C. $\\dfrac{a+2b}{3}$","D. $\\dfrac{a}{3}$"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{{{a}^{2}}-2ab}{{{a}^{2}}b}\\cdot P=\\dfrac{{{a}^{2}}b-4{{b}^{3}}}{3a{{b}^{2}}} \\\\ & \\Rightarrow P=\\dfrac{{{a}^{2}}b-4{{b}^{3}}}{3a{{b}^{2}}}:\\dfrac{{{a}^{2}}-2ab}{{{a}^{2}}b} \\\\ & =\\dfrac{b\\left( {{a}^{2}}-4{{b}^{2}} \\right)}{3a{{b}^{2}}}:\\dfrac{a\\left( a-2b \\right)}{{{a}^{2}}b} \\\\ & =\\dfrac{b\\left( a+2b \\right)\\left( a-2b \\right)}{3a{{b}^{2}}}\\cdot \\dfrac{{{a}^{2}}b}{a\\left( a-2b \\right)} \\\\ & =\\dfrac{a+2b}{3} \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span>","column":2}]}],"id_ques":131},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2016"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv2/img\/12.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{{{x}^{2}}-4}{9-{{y}^{2}}}:\\dfrac{x-2}{3+y}-\\dfrac{2}{3-y}$ t\u1ea1i $x = 2016;y=2$ l\u00e0 _input_ ","hint":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau \u0111\u00f3 thay $x = 2016;y=2$ 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\/> $\\begin{align} & \\dfrac{{{x}^{2}}-4}{9-{{y}^{2}}}:\\dfrac{x-2}{3+y}-\\dfrac{2}{3-y} \\\\ & =\\dfrac{\\left( x+2 \\right)\\left( x-2 \\right)}{\\left( 3+y \\right)\\left( 3-y \\right)}\\cdot \\dfrac{3+y}{x-2}-\\dfrac{2}{3-y} \\\\ & =\\dfrac{x+2}{3-y}-\\dfrac{2}{3-y} \\\\ & =\\dfrac{x+2-2}{3-y} \\\\ & =\\dfrac{x}{3-y} \\\\ \\end{align}$ <br\/> Thay $x = 2016;y=2$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{x}{3-y}=\\dfrac{2016}{3-2}=2016$<\/span> "}]}],"id_ques":132},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["36"],["25"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv2/img\/11.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{x+1}{x+2}:\\left( \\dfrac{x+2}{x+3}:\\dfrac{x+3}{x+1} \\right)$ t\u1ea1i $x = 3$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{x+1}{x+2}:\\left( \\dfrac{x+2}{x+3}:\\dfrac{x+3}{x+1} \\right) \\\\ &=\\dfrac{x+1}{x+2}:\\left( \\dfrac{x+2}{x+3}\\cdot \\dfrac{x+1}{x+3} \\right) \\\\ & =\\dfrac{x+1}{x+2}:\\dfrac{\\left( x+2 \\right)\\left( x+1 \\right)}{{{\\left( x+3 \\right)}^{2}}} \\\\ & =\\dfrac{x+1}{x+2}\\cdot \\dfrac{{{\\left( x+3 \\right)}^{2}}}{\\left( x+2 \\right)\\left( x+1 \\right)} \\\\ & =\\dfrac{\\left( x+1 \\right){{\\left( x+3 \\right)}^{2}}}{{{\\left( x+2 \\right)}^{2}}\\left( x+1 \\right)} \\\\ & ={{\\left( \\dfrac{x+3}{x+2} \\right)}^{2}} \\\\ \\end{align}$ <br\/> Thay $x = 3$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> ${{\\left( \\dfrac{x+3}{x+2} \\right)}^{2}}={{\\left( \\dfrac{3+3}{3+2} \\right)}^{2}}=\\dfrac{36}{25}$ <\/span> "}]}],"id_ques":133},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-1"],["6"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv2/img\/10.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{a-b}{{{a}^{2}}+ab+a+b}:\\dfrac{{{b}^{2}}-ab+b-a}{a+b}$ t\u1ea1i $a=1; b=2$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} &\\dfrac{a-b}{{{a}^{2}}+ab+a+b}:\\dfrac{{{b}^{2}}-ab+b-a}{a+b} \\\\ &=\\dfrac{a-b}{a\\left( a+b \\right)+\\left( a+b \\right)}:\\dfrac{b\\left( b-a \\right)+\\left( b-a \\right)}{a+b} \\\\ & =\\dfrac{\\left( a-b \\right)}{\\left( a+b \\right)\\left( a+1 \\right)}:\\dfrac{\\left( b-a \\right)\\left( b+1 \\right)}{a+b} \\\\ & =\\dfrac{\\left( a-b \\right)}{\\left( a+b \\right)\\left( a+1 \\right)}\\cdot \\dfrac{\\left( a+b \\right)}{\\left( b-a \\right)\\left( b+1 \\right)} \\\\ & =\\dfrac{\\left( a-b \\right)\\left( a+b \\right)}{\\left( a+b \\right)\\left( a+1 \\right)\\left( b-a \\right)\\left( b+1 \\right)} \\\\ & =\\dfrac{-1}{\\left( a+1 \\right)\\left( b+1 \\right)} \\\\ \\end{align}$ <br\/> Thay $a=1; b=2$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{-1}{\\left( a+1 \\right)\\left( b+1 \\right)}=\\dfrac{-1}{\\left( 1+1 \\right)\\left( 2+1 \\right)}=\\dfrac{-1}{6}$<\/span> "}]}],"id_ques":134},{"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/lv2/img\/9.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{2a\\,x-4ay}{{{x}^{2}}+4xy+4{{y}^{2}}}:\\dfrac{4a{{y}^{2}}-a\\,{{x}^{2}}}{{{x}^{2}}-4xy+4{{y}^{2}}}$ l\u00e0: ","select":["A. $\\dfrac{2}{x+2y}$ ","B. $\\dfrac{2(x+2y)}{(x-2y)^2}$","C. $\\dfrac{-2(x-2y)^2}{(x+2y)^3}$","D. $\\dfrac{x-2y}{x+2y}$"],"explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} &\\dfrac{2a\\,x-4ay}{{{x}^{2}}+4xy+4{{y}^{2}}}:\\dfrac{4a{{y}^{2}}-a\\,{{x}^{2}}}{{{x}^{2}}-4xy+4{{y}^{2}}} \\\\ &=\\dfrac{2a\\left( x-2y \\right)}{{{\\left( x+2y \\right)}^{2}}}:\\dfrac{a\\left( 4{{y}^{2}}-{{x}^{2}} \\right)}{{{\\left( x-2y \\right)}^{2}}} \\\\ & =\\dfrac{2a\\left( x-2y \\right)}{{{\\left( x+2y \\right)}^{2}}}\\cdot \\dfrac{{{\\left( x-2y \\right)}^{2}}}{a\\left( 2y+x \\right)\\left( 2y-x \\right)} \\\\ & =\\dfrac{2a{{\\left( x-2y \\right)}^{3}}}{a{{\\left( x+2y \\right)}^{3}}\\left( 2y-x \\right)} \\\\ & =\\dfrac{-2{{\\left( x-2y \\right)}^{2}}}{{{\\left( x+2y \\right)}^{3}}} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span><\/span> ","column":2}]}],"id_ques":135},{"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/lv2/img\/8.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{{{x}^{2}}+x}{5{{x}^{2}}-10x+5}:\\dfrac{3x+3}{5x-5}$ l\u00e0: ","select":["A. $\\dfrac{1}{x-1}$ ","B. $\\dfrac{x-1}{x}$","C. $\\dfrac{x}{3(x-1)}$","D. $\\dfrac{x}{x-1}$"],"explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{{{x}^{2}}+x}{5{{x}^{2}}-10x+5}:\\dfrac{3x+3}{5x-5} \\\\ &=\\dfrac{x\\left( x+1 \\right)}{5\\left( {{x}^{2}}-2x+1 \\right)}:\\dfrac{3\\left( x+1 \\right)}{5\\left( x-1 \\right)} \\\\ & =\\dfrac{x\\left( x+1 \\right)}{5{{\\left( x-1 \\right)}^{2}}}\\cdot \\dfrac{5\\left( x-1 \\right)}{3\\left( x+1 \\right)} \\\\ & =\\dfrac{x\\left( x+1 \\right)5\\left( x-1 \\right)}{5{{\\left( x-1 \\right)}^{2}}3\\left( x+1 \\right)} \\\\ & =\\dfrac{x}{3\\left( x-1 \\right)} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span><\/span> ","column":2}]}],"id_ques":136},{"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/lv2/img\/5.jpg' \/><\/center> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\left( x-\\dfrac{{{x}^{2}}+{{y}^{2}}}{x+y} \\right)\\cdot \\left( \\dfrac{1}{y}+\\dfrac{2}{x-y} \\right)$ l\u00e0: ","select":["A. $1$ ","B. $x-y$","C. $\\dfrac{1}{x+y}$","D. $\\dfrac{1}{x}$"],"explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\left( x-\\dfrac{{{x}^{2}}+{{y}^{2}}}{x+y} \\right)\\cdot \\left( \\dfrac{1}{y}+\\dfrac{2}{x-y} \\right)$<br\/> $=\\left( \\dfrac{x\\left( x+y \\right)}{x+y}-\\dfrac{{{x}^{2}}+{{y}^{2}}}{x+y} \\right)$ $\\cdot \\left( \\dfrac{x-y}{y\\left( x-y \\right)}+\\dfrac{2y}{y\\left( x-y \\right)} \\right) $<br\/> $ =\\dfrac{x\\left( x+y \\right)-{{x}^{2}}-{{y}^{2}}}{x+y}$ $\\cdot \\dfrac{x-y+2y}{y\\left( x-y \\right)} $<br\/> $ =\\dfrac{{{x}^{2}}+xy-{{x}^{2}}-{{y}^{2}}}{x+y}\\cdot \\dfrac{x+y}{y\\left( x-y \\right)} $<br\/> $ =\\dfrac{xy-{{y}^{2}}}{x+y}\\cdot \\dfrac{x+y}{y\\left( x-y \\right)} $<br\/> $ =\\dfrac{y\\left( x-y \\right)\\left( x+y \\right)}{\\left( x+y \\right)y\\left( x-y \\right)}$<br\/> $ =1$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":2}]}],"id_ques":137},{"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":[[["1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv2/img\/4.jpg' \/><\/center> $\\dfrac{34xy}{12{{x}^{2}}{{y}^{2}}}\\cdot \\dfrac{6xyz}{17z}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ ","hint":" <b> Nh\u00e2n hai ph\u00e2n th\u1ee9c:<\/b> <br\/> $\\dfrac{A}{B} \\cdot \\dfrac{C}{D}=\\dfrac{A.C}{B.D}$ v\u1edbi $B, D \\ne 0$","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} &\\dfrac{34xy}{12{{x}^{2}}{{y}^{2}}}\\cdot \\dfrac{6xyz}{17z} \\\\ &=\\dfrac{34}{12xy}\\cdot \\dfrac{6xy}{17} \\\\ & =\\dfrac{17.6xy}{6xy.17} \\\\ & =1 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $1$ <\/span><\/span> "}]}],"id_ques":138},{"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+6","6+x"],["x+5","5+x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv2/img\/3.jpg' \/><\/center> $\\dfrac{{{x}^{2}}-36}{2x+10}\\cdot \\dfrac{3}{6-x}=\\dfrac{-3.(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})}{2.(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} &\\dfrac{{{x}^{2}}-36}{2x+10}\\cdot \\dfrac{3}{6-x} \\\\ &=\\dfrac{\\left( x+6 \\right)\\left( x-6 \\right)}{2\\left( x+5 \\right)}\\cdot \\dfrac{3}{6-x} \\\\ & =\\dfrac{\\left( x+6 \\right)\\left( x-6 \\right)3}{2\\left( x+5 \\right)\\left( 6-x \\right)} \\\\ & =\\dfrac{-3\\left( x+6 \\right)}{2\\left( x+5 \\right)} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{-3\\left( x+6 \\right)}{2\\left( x+5 \\right)}$ <\/span><\/span> "}]}],"id_ques":139},{"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":[[["x"],["x-2"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai14/lv2/img\/2.jpg' \/><\/center> $\\dfrac{{{x}^{3}}-8}{5x+20}\\cdot \\dfrac{{{x}^{2}}+4x}{{{x}^{2}}+2x+4}$ $=\\dfrac{(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}).(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})}{5}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & \\dfrac{{{x}^{3}}-8}{5x+20}\\cdot \\dfrac{{{x}^{2}}+4x}{{{x}^{2}}+2x+4}\\\\ &=\\dfrac{\\left( x-2 \\right)\\left( {{x}^{2}}+2x+4 \\right)}{5\\left( x+4 \\right)}\\cdot \\dfrac{x\\left( x+4 \\right)}{{{x}^{2}}+2x+4} \\\\ & =\\dfrac{\\left( x-2 \\right)\\left( {{x}^{2}}+2x+4 \\right)x\\left( x+4 \\right)}{5\\left( x+4 \\right)\\left( {{x}^{2}}+2x+4 \\right)} \\\\ & =\\dfrac{x\\left( x-2 \\right)}{5} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{x\\left( x-2 \\right)}{5}$ <\/span><\/span> "}]}],"id_ques":140}],"lesson":{"save":0,"level":2}}