{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t cho c\u00e2u r\u00fat g\u1ecdn ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai12/lv3/img\/12.jpg' \/><\/center> $\\dfrac{a^4-3a^2+1}{a^4-a^2-2a-1}$$=\\dfrac{(a^4-3a^2+1):\\square }{(a^4-a^2-2a-1):\\square }$$=\\dfrac{a^2+a-1}{a^2+a+1}$<br\/> Bi\u1ec3u th\u1ee9c c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c ph\u00e9p r\u00fat g\u1ecdn \u0111\u00fang l\u00e0: ","select":["A. $a^2-a-1$ ","B. $a(a-1)$ ","C. $a^2-1$","D. $a(a+1)$"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $ \\dfrac{{{a}^{4}}-3{{a}^{2}}+1}{{{a}^{4}}-{{a}^{2}}-2a-1}$<br\/>$=\\dfrac{{{a}^{4}}-2{{a}^{2}}+1-{{a}^{2}}}{{{a}^{4}}-\\left( {{a}^{2}}+2a+1 \\right)} $<br\/>$ =\\dfrac{{{\\left( {{a}^{2}}-1 \\right)}^{2}}-{{a}^{2}}}{{{a}^{4}}-{{\\left( a+1 \\right)}^{2}}} $<br\/>$ =\\dfrac{\\left( {{a}^{2}}-1+a \\right)\\left( {{a}^{2}}-1-a \\right)}{\\left( {{a}^{2}}+a+1 \\right)\\left( {{a}^{2}}-a-1 \\right)} $<br\/>$ =\\dfrac{\\left( {{a}^{2}}+a-1 \\right)\\left( {{a}^{2}}-a-1 \\right):\\left( {{a}^{2}}-a-1 \\right)}{\\left( {{a}^{2}}+a+1 \\right)\\left( {{a}^{2}}-a-1 \\right):\\left( {{a}^{2}}-a-1 \\right)} $<br\/>$ =\\dfrac{{{a}^{2}}+a-1}{{{a}^{2}}+a+1} $<br\/> C\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u00e0 $a^2-a-1$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":2}]}],"id_ques":41},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai12/lv3/img\/9.jpg' \/><\/center>Cho $\\dfrac{x}{a}=\\dfrac{y}{b}=\\dfrac{z}{c}\\ne 0$ <br\/> R\u00fat g\u1ecdn ph\u00e2n th\u1ee9c $\\dfrac{(x^2+y^2+z^2)(a^2+b^2+c^2)}{(ax+by+cz)^2}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $1$ ","B. $a+b+c$ ","C. $x+y+z$","D. $a^2+b^2+c^2$"],"hint":"\u0110\u1eb7t $\\dfrac{x}{a}=\\dfrac{y}{b}=\\dfrac{z}{c}=k\\ne 0$ <br\/> R\u00fat $x, y, z$ theo $k, a, b, c$. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> \u0110\u1eb7t $\\dfrac{x}{a}=\\dfrac{y}{b}=\\dfrac{z}{c}=k\\ne 0$<br\/> <b> B\u01b0\u1edbc 2:<\/b> R\u00fat $x, y, z$ theo $k, a, b, c$. Thay $x, y, z$ \u0111\u00f3 v\u00e0o ph\u00e2n th\u1ee9c.<br\/> <b> B\u01b0\u1edbc 3:<\/b> Ti\u1ebfp t\u1ee5c r\u00fat g\u1ecdn ph\u00e2n th\u1ee9c. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> \u0110\u1eb7t $\\dfrac{x}{a}=\\dfrac{y}{b}=\\dfrac{z}{c}=k\\ne 0$ th\u00ec $x=ka; y=kb; z=kc$.<br\/> Thay $x=ka; y=kb; z=kc$ v\u00e0o ph\u00e2n th\u1ee9c, ta \u0111\u01b0\u1ee3c:<br\/> $\\begin{align} &\\dfrac{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}{{{(ax+by+cz)}^{2}}}\\\\&=\\dfrac{\\left( {{k}^{2}}{{a}^{2}}+{{k}^{2}}{{b}^{2}}+{{k}^{2}}{{c}^{2}} \\right)\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right)}{{{\\left( k{{a}^{2}}+k{{b}^{2}}+k{{c}^{2}} \\right)}^{2}}} \\\\ & =\\dfrac{{{k}^{2}}\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right)\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right)}{{{k}^{2}}{{\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right)}^{2}}} \\\\ & =\\dfrac{{{k}^{2}}{{\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right)}^{2}}}{{{k}^{2}}{{\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right)}^{2}}} \\\\ & =1 \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":42},{"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/bai12/lv3/img\/8.jpg' \/><\/center> R\u00fat g\u1ecdn ph\u00e2n th\u1ee9c $\\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $\\dfrac{1}{x+y}$ ","B. $\\dfrac{x-y}{x+y}$ ","C. $\\dfrac{x+y}{x-y}$","D. $\\dfrac{1}{x-y}$"],"explain":"<span class='basic_left'>Ta c\u00f3: <br\/> $\\begin{align} &\\dfrac{{{x}^{2}}+3xy+2{{y}^{2}}}{{{x}^{3}}+2{{x}^{2}}y-x{{y}^{2}}-2{{y}^{3}}}\\\\&=\\dfrac{{{x}^{2}}+xy+2xy+2{{y}^{2}}}{{{x}^{3}}-x{{y}^{2}}+2{{x}^{2}}y-2{{y}^{3}}} \\\\ & =\\dfrac{x\\left( x+y \\right)+2y\\left( x+y \\right)}{x\\left( {{x}^{2}}-{{y}^{2}} \\right)+2y\\left( {{x}^{2}}-{{y}^{2}} \\right)} \\\\ & =\\dfrac{\\left( x+y \\right)\\left( x+2y \\right)}{\\left( {{x}^{2}}-{{y}^{2}} \\right)\\left( x+2y \\right)} \\\\ & =\\dfrac{x+y}{\\left( x+y \\right)\\left( x-y \\right)} \\\\ & =\\dfrac{1}{x-y} \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":43},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai12/lv3/img\/4.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a ph\u00e2n th\u1ee9c $\\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn $x, y$ ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> R\u00fat g\u1ecdn ph\u00e2n th\u1ee9c $\\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}$ <br\/> N\u1ebfu ph\u00e2n th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn kh\u00f4ng ch\u1ee9a bi\u1ebfn th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a ph\u00e2n th\u1ee9c kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn v\u00e0 ng\u01b0\u1ee3c l\u1ea1i.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> Ta c\u00f3: <br\/> $\\dfrac{{{x}^{3}}}{{{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}}$$=\\dfrac{{{x}^{3}}}{{{\\left( x-y \\right)}^{3}}}$ <br\/> V\u1eady gi\u00e1 tr\u1ecb ph\u00e2n th\u1ee9c ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn $x, y$. <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":44},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-2"],["3"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai12/lv3/img\/2.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a ph\u00e2n th\u1ee9c $\\dfrac{0,5x^2+x+2}{x^4-8x}$ t\u1ea1i $x =\\dfrac{1}{2}$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{0,5{{x}^{2}}+x+2}{{{x}^{4}}-8x} \\\\ & =\\dfrac{0,5\\left( {{x}^{2}}+2x+4 \\right)}{x\\left( {{x}^{3}}-8 \\right)} \\\\ & =\\dfrac{0,5\\left( {{x}^{2}}+2x+4 \\right)}{x\\left( x-2 \\right)\\left( {{x}^{2}}+2x+4 \\right)} \\\\ & =\\dfrac{0,5}{x\\left( x-2 \\right)} \\\\ \\end{align}$<br\/> Thay $x=\\dfrac{1}{2}$ v\u00e0o ph\u00e2n th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{0,5}{x\\left( x-2 \\right)}=\\dfrac{0,5}{\\dfrac{1}{2}\\left( \\dfrac{1}{2}-2 \\right)}$$=\\dfrac{\\dfrac{1}{2}}{\\dfrac{1}{2}.\\left( -\\dfrac{3}{2} \\right)}$$=\\dfrac{-2}{3}$<\/span> "}]}],"id_ques":45},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-8"],["3"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai12/lv3/img\/13.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a ph\u00e2n th\u1ee9c $\\dfrac{ax^4-a^4x}{a^2+ax+x^2}$ t\u1ea1i $a=3;x = \\dfrac{1}{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{ax^4-a^4x}{a^2+ax+x^2} \\\\ & =\\dfrac{ax(x^3-a^3)}{a^2+ax+x^2} \\\\ & =\\dfrac{ax(x-a)(x^2+ax+a^2)}{a^2+ax+x^2} \\\\ & =ax(x-a) \\\\ \\end{align}$ <br\/> Thay $a=3;x = \\dfrac{1}{3}$ v\u00e0o ph\u00e2n th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c: <br\/> $ax(x-a)=3.\\dfrac{1}{3}(\\dfrac{1}{3}-3)=\\dfrac{-8}{3}$<\/span> "}]}],"id_ques":46},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-1"],["8"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai12/lv3/img\/12.jpg' \/><\/center> Ph\u00e2n th\u1ee9c $\\dfrac{xy}{x^2-7x-8}$ c\u00f3 ngh\u0129a khi $\\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.$ ","explain":"<span class='basic_left'> Ph\u00e2n th\u1ee9c $\\dfrac{xy}{x^2-7x-8}$ c\u00f3 ngh\u0129a khi <br\/> $\\begin{aligned} & x^2-7x-8\\,\\,\\ne 0 \\\\ &\\Rightarrow x^2-8x+x-8\\,\\,\\ne 0 \\\\ & \\Rightarrow x(x-8)+(x-8)\\,\\,\\ne 0 \\\\ &\\Rightarrow (x-8)\\left( x+1 \\right)\\,\\,\\ne 0 \\\\ & \\Rightarrow \\left\\{ \\begin{aligned} & x-8\\ne 0 \\\\ & x+1\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 8 \\\\ & x\\ne -1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $8$ v\u00e0 $-1$. <\/span><\/span> "}]}],"id_ques":47},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai12/lv3/img\/9.jpg' \/><\/center> $\\dfrac{3x^2-12x+12}{x^4-8x}$ v\u00e0 $\\dfrac{3(x-2)}{x(x^2+2x+4)}$ l\u00e0 hai ph\u00e2n th\u1ee9c b\u1eb1ng nhau. ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\dfrac{3x^2-12x+12}{x^4-8x}\\\\ =\\dfrac{3(x^2-4x+4)}{x(x^3-8)}$ <br\/> $=\\dfrac{3(x-2)^2}{x(x-2)(x^2+2x+4)}$ <br\/> $=\\dfrac{3(x-2)}{x(x^2+2x+4)}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":48},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong c\u00e2u r\u00fat g\u1ecdn ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["2x-y","-y+2x"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai12/lv3/img\/4.jpg' \/><\/center> $\\dfrac{x-y}{(x-y)^2+x(x-y)}$$=\\dfrac{1}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\dfrac{x-y}{(x-y)^2+x(x-y)}$ <br\/> $=\\dfrac{x-y}{(x-y)(x-y+x)}$ <br\/> $=\\dfrac{x-y}{(x-y)(2x-y)}$ <br\/> $=\\dfrac{1}{2x-y} $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2x-y$. <\/span><\/span> "}]}],"id_ques":49},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong c\u00e2u r\u00fat g\u1ecdn ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["2x-1","-1+2x"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai12/lv3/img\/2.jpg' \/><\/center> $\\dfrac{2x^2+5x-3}{x^2-9}=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{x-3}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\dfrac{2x^2+5x-3}{x^2-9}$ <br\/> $=\\dfrac{2x^2+6x-x-3}{(x+3)(x-3)}$ <br\/> $=\\dfrac{2x(x+3)-(x+3)}{(x+3)(x-3)} $ <br\/> $=\\dfrac{(x+3)(2x-1)}{(x+3)(x-3)}$ <br\/> $=\\dfrac{2x-1}{x-3}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2x-1$. <\/span><\/span> "}]}],"id_ques":50}],"lesson":{"save":0,"level":3}}