{"segment":[{"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/bai3/lv3/img\/3.jpg' \/><\/center> R\u00fat g\u1ecdn $4{{x}^{2}}+2{{z}^{2}}-4xz-2z+1 $ ta \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $(2x-z)^2+(z-1)^2$","B. $(2x-z)^2+z^2$","C. $(x-2z)^2+(z+1)^2$","D. $(2x-1)^2+(z-1)^2$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ta nh\u00f3m: $ 4x^2-4xz+z^2$ v\u00e0 nh\u00f3m: $z^2-2z+1$.<br\/> <b> B\u01b0\u1edbc 2: <\/b> \u0110\u01b0a bi\u1ec3u th\u1ee9c v\u1ec1 c\u00e1c b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t hi\u1ec7u.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & 4{{x}^{2}}+2{{z}^{2}}-4xz-2z+1 \\\\ & =\\left( 4{{x}^{2}}-4xz+{{z}^{2}} \\right)+\\left( {{z}^{2}}-2z+1 \\right) \\\\ & ={{\\left( 2x-z \\right)}^{2}}+{{\\left( z-1 \\right)}^{2}} \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":421},{"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/bai3/lv3/img\/3.jpg' \/><\/center> Gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c $M=\\left( {{50}^{2}}+{{48}^{2}}+...+{{2}^{2}} \\right)-\\left( {{49}^{2}}+{{47}^{2}}+...+1 \\right)$ l\u00e0:","select":["A. $1375$","B. $1275$","C. $1475$","D. $1575$"],"hint":"Ta ph\u00e1 ngo\u1eb7c v\u00e0 nh\u00f3m th\u00e0nh c\u00e1c nh\u00f3m l\u00e0 hi\u1ec7u b\u00ecnh ph\u01b0\u01a1ng hai s\u1ed1 li\u00ean ti\u1ebfp. <br\/> T\u1eeb \u0111\u00f3 \u00e1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng \u0111\u1ec3 khai tri\u1ec3n v\u00e0 t\u00ednh to\u00e1n.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ph\u00e1 ngo\u1eb7c <br\/> <b> B\u01b0\u1edbc 2: <\/b> Nh\u00f3m c\u00e1c c\u1eb7p s\u1ed1 \u0111\u1ec3 \u0111\u01b0\u1ee3c c\u00e1c nh\u00f3m l\u00e0 hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng <br\/> <b> B\u01b0\u1edbc 3:<\/b> \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng \u0111\u1ec3 khai tri\u1ec3n <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $ M=\\left( {{50}^{2}}+{{48}^{2}}+...+{{2}^{2}} \\right)-\\left( {{49}^{2}}+{{47}^{2}}+...+1 \\right) $<br\/>$ =\\left( {{50}^{2}}-{{49}^{2}} \\right)+\\left( {{48}^{2}}-{{47}^{2}} \\right)+...+\\left( {{2}^{2}}-1 \\right) $<br\/>$ =\\left( 50+49 \\right)\\left( 50-49 \\right)+\\left( 48+47 \\right)\\left( 48-47 \\right)+...+\\left( 2+1 \\right)\\left( 2-1 \\right) $<br\/>$ =99.1+95.1+...+3.1 $<br\/>$ =99+95+...+3 $<br\/>$ =\\dfrac{\\left[ \\left( 99-3 \\right):4+1 \\right]}{2}.\\left( 99+3 \\right) $<br\/>$ =1275 $ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":422},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai3/lv3/img\/3.jpg' \/><\/center>Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $A=x^2+10y^2-6xy-2y+3$ l\u00e0 _input_","hint":"\u0110\u01b0a bi\u1ec3u th\u1ee9c $A$ v\u1ec1 d\u1ea1ng: $A=[f(x,y)]^2+ [f(y)]^2+a$ v\u1edbi $a$ l\u00e0 h\u1eb1ng s\u1ed1 <br\/>Sau \u0111\u00f3 t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a A.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> \u0110\u01b0a bi\u1ec3u th\u1ee9c $A$ v\u1ec1 d\u1ea1ng: $A=[f(x,y)]^2+ [f(y)]^2+[f(x)]^2+a$ <br\/> V\u1edbi $a$ l\u00e0 h\u1eb1ng s\u1ed1; $f(x,y)$ l\u00e0 bi\u1ec3u th\u1ee9c ch\u1ee9a bi\u1ebfn $x, y; f(y)$ l\u00e0 bi\u1ec3u th\u1ee9c ch\u1ee9a bi\u1ebfn $y$. <br\/> <b> B\u01b0\u1edbc 2: <\/b> Sau \u0111\u00f3 \u0111\u00e1nh gi\u00e1: $A=[f(x,y)]^2+ [f(y)]^2+[f(x)]^2+a\\ge a$,<br\/> <b> B\u01b0\u1edbc 3: <\/b> K\u1ebft lu\u1eadn gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $A$, \u0111\u1ea1t \u0111\u01b0\u1ee3c khi n\u00e0o. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $\\begin{aligned} & A={{x}^{2}}+10{{y}^{2}}-6xy-2y+3 \\\\ & =\\left( {{x}^{2}}-6xy+9{{y}^{2}} \\right)+\\left( {{y}^{2}}-2y+1 \\right)+2 \\\\ & ={{\\left( x-3y \\right)}^{2}}+{{\\left( y-1 \\right)}^{2}}+2 \\\\ & Do\\,\\,\\,\\left\\{ \\begin{aligned} & {{\\left( x-3y \\right)}^{2}}\\,\\,\\ge \\,\\,0 \\\\ & {{\\left( y-1 \\right)}^{2}}\\,\\,\\ge \\,\\,0, (\\forall x) \\\\ \\end{aligned} \\right.\\Rightarrow \\,\\,A\\,\\,\\,\\ge \\,\\,2 \\\\ \\end{aligned}$<br\/> V\u1eady GTNN c\u1ee7a $A$ l\u00e0 $2$ <br\/> $A$ \u0111\u1ea1t GTNN khi v\u00e0 ch\u1ec9 khi $\\left\\{ \\begin{aligned} & x-3y=0 \\\\ & y-1=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x=3 \\\\ & y=1 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$.<\/span>"}]}],"id_ques":423},{"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/bai3/lv3/img\/4.jpg' \/><\/center>Bi\u1ec3u th\u1ee9c $\\left( 7n-2 \\right)^2-\\left( 2n-7 \\right)^2$ lu\u00f4n chia h\u1ebft cho 9 v\u1edbi m\u1ecdi n $\\in\\mathbb{Z}$, <b> \u0111\u00fang <\/b> hay <b> sai <\/b>? ","select":["\u0110\u00fang","Sai"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ta bi\u1ebfn \u0111\u1ed5i v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c, s\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c H\u0110T th\u1ee9 ba: $a^2-b^2=(a+b)(a-b)$. <br\/> <b> B\u01b0\u1edbc 2: <\/b> Nh\u1eadn \u0111\u1ecbnh k\u1ebft qu\u1ea3 \u0111\u00e3 r\u00fat g\u1ecdn xem c\u00f3 ch\u1ee9a th\u1eeba s\u1ed1 chia h\u1ebft cho 9 kh\u00f4ng.<br\/> + N\u1ebfu c\u00f3 ch\u1ee9a th\u1eeba s\u1ed1 chia h\u1ebft cho 9 th\u00ec bi\u1ec3u th\u1ee9c \u0111\u00e3 cho chia h\u1ebft cho $9$.<br\/> + N\u1ebfu kh\u00f4ng ch\u1ee9a th\u1eeba s\u1ed1 chia h\u1ebft cho $9$ th\u00ec bi\u1ec3u th\u1ee9c \u0111\u00e3 cho kh\u00f4ng chia h\u1ebft cho $9$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c:<br\/>$ {{\\left( 7n-2 \\right)}^{2}}-{{\\left( 2n-7 \\right)}^{2}} $<br\/>$ =\\left( 7n-2+2n-7 \\right)\\left( 7n-2-2n+7 \\right) $<br\/>$ =\\left( 9n-9 \\right)\\left( 5n+5 \\right) $<br\/>$ =9\\left( n-1 \\right).5\\left( n+1 \\right) $<br\/>$ =45\\left( n-1 \\right)\\left( n+1 \\right) $<br\/>$ =45\\left( {{n}^{2}}-1 \\right) $ <br\/> Do $45\\vdots 9 \\Rightarrow 45(n^2-1) \\vdots 9$ v\u1edbi m\u1ecdi $n \\in \\mathbb{Z}$<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}]}],"id_ques":424},{"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":"Bi\u1ebft $(2x+3)^2-(2x+1)(2x-1)=22$, gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0:","select":["A. $0$","B. $1$","C. $2$","D. $3$"],"hint":"Bi\u1ebfn \u0111\u1ed5i v\u1ebf tr\u00e1i, s\u1eed d\u1ee5ng c\u00e1c h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $(a+b)^2=a^2+2ab+b^2$, $ a^2-b^2=(a+b)(a-b)$. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Bi\u1ebfn \u0111\u1ed5i v\u1ebf tr\u00e1i, s\u1eed d\u1ee5ng c\u00e1c h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $(a+b)^2=a^2+2ab+b^2$, $ a^2-b^2=(a+b)(a-b)$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $ {{(2x+3)}^{2}}-(2x+1)(2x-1) =22 \\\\ \\Leftrightarrow {{(2x+3)}^{2}}-\\left( 4{{x}^{2}}-1 \\right) =22 \\\\ \\Leftrightarrow 4{{x}^{2}}+12x+9-4{{x}^{2}}+1=22 \\\\ \\Leftrightarrow 12x+10 =22 \\\\ \\Leftrightarrow 12x=12 \\\\ \\Leftrightarrow x =1 $ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span><\/span>","column":2}]}],"id_ques":425},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai3/lv3/img\/1.png' \/><\/center>Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $A={{\\left( 2x+3 \\right)}^{2}}+\\left( 2x+3 \\right)\\left( 2x-6 \\right)+{{\\left( x-3 \\right)}^{2}} $ t\u1ea1i $x = \\dfrac{3}{4}$ l\u00e0: ","select":["A. $\\dfrac{25}{4}$","B. $\\dfrac{16}{9}$","C. $\\dfrac{81}{16}$","D. $\\dfrac{81}{9}$"],"hint":"Coi $a = 2x+3; b = x-3$ th\u00ec bi\u1ec3u th\u1ee9c $A$ l\u00e0 khai tri\u1ec3n c\u1ee7a h\u1eb1ng \u0111\u1eb3ng th\u1ee9c n\u00e0o?","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ta l\u00e0m g\u1ecdn bi\u1ec3u th\u1ee9c $A$, s\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $(a+b)^2=a^2+2ab+b^2$, coi $a = 2x+3; b = x-3$<br\/> <b> B\u01b0\u1edbc 2: <\/b> Thay $x = \\dfrac{3}{4}$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn v\u00e0 t\u00ednh.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $A ={{\\left( 2x+3 \\right)}^{2}}+\\left( 2x+3 \\right)\\left( 2x-6 \\right)+{{\\left( x-3 \\right)}^{2}} $<br\/>$ ={{\\left( 2x+3 \\right)}^{2}}+\\left( 2x-6 \\right)\\left( 2x+3 \\right)+{{\\left( x-3 \\right)}^{2}} $<br\/>$ ={{\\left( 2x+3 \\right)}^{2}}+2\\left( x-3 \\right)\\left( 2x+3 \\right)+{{\\left( x-3 \\right)}^{2}} $<br\/>$ ={{\\left[ \\left( 2x+3 \\right)+\\left( x-3 \\right) \\right]}^{2}} $<br\/>$ ={{\\left( 3x \\right)}^{2}} $<br\/>$ =9{{x}^{2}} $. <br\/>Thay $x = \\dfrac{3}{4}$ v\u00e0o $A$, ta \u0111\u01b0\u1ee3c: <br\/>$A=9.{{\\left( \\dfrac{3}{4} \\right)}^{2}}=9.\\dfrac{9}{16}=\\dfrac{81}{16}$ . <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span>","column":2}]}],"id_ques":426},{"time":24,"part":[{"title":"\u0110i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c khai tri\u1ec3n \u0111\u00fang","title_trans":"","temp":"fill_the_blank","correct":[[["2"],["15"],["4"],["225"],["1"],["9"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["frac","sqr"],"ques":"$\\left( \\dfrac{2}{15}{{x}^{2}}y+\\dfrac{1}{3}yz \\right)$ (<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>${{x}^{2}}y$ -$\\dfrac{1}{3}yz)$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>${{x}^{4}}y^2$ - <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>${{y}^{2}}{{z}^{2}}$ ","hint":"Khai tri\u1ec3n tr\u00ean l\u00e0 d\u1ea1ng c\u1ee7a h\u1eb1ng \u0111\u1eb3ng th\u1ee9c n\u00e0o? ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u1eeb $\\left( \\dfrac{2}{15}{{x}^{2}}y+\\dfrac{1}{3}yz \\right)$ suy ra $a, b$. <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u1eeb $a$ v\u00e0 $b$ \u0111\u00e3 t\u00ecm suy ra h\u1ec7 s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\left( \\dfrac{2}{15}{{x}^{2}}y+\\dfrac{1}{3}yz \\right)$ suy ra $ a =\\dfrac{2}{15}{{x}^{2}}y$; $b = \\dfrac{1}{3}yz$. <br\/> Suy ra $ a^2 =\\dfrac{4}{225}{{x}^{4}}y^2$; $b^2 = \\dfrac{1}{9}y^2z^2$. <br\/> Ta c\u00f3 khai tri\u1ec3n ho\u00e0n ch\u1ec9nh l\u00e0: $\\left( \\dfrac{2}{15}{{x}^{2}}y+\\dfrac{1}{3}yz \\right)\\left( \\dfrac{2}{15}{{x}^{2}}y-\\dfrac{1}{3}yz \\right)$$=\\dfrac{4}{225}{{x}^{4}}{{y}^{2}}-\\dfrac{1}{9}{{y}^{2}}{{z}^{2}}$ ."}]}],"id_ques":427},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $x^2; x$","B. $x^2; 2x$","C. $x^2; x^2$"],"ques":"?$+40x+400=$(?$+20)^2$","hint":"Ta th\u1ea5y \u0111\u01b0\u1ee3c \u0111\u00e2y l\u00e0 khai tri\u1ec3n h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $a^2+2ab+b^2=(a+b)^2$<br\/> Ta \u0111i t\u00ecm c\u00e1c \u00f4 c\u1ea7n \u0111i\u1ec1n d\u1ef1a v\u00e0o c\u00e1c y\u1ebfu t\u1ed1 \u0111\u00e3 cho trong khai tri\u1ec3n.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> X\u00e1c \u0111\u1ecbnh $b^2$ trong khai tri\u1ec3n H\u0110T.<br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u1eeb $b^2$ suy ra $b$. <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u1eeb $2ab$ \u0111\u1ec3 t\u00ecm ra $a$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $b^2=400 \\Rightarrow b = 20$.<br\/> $2ab=40x \\Leftrightarrow 2.x.20=40x \\Rightarrow a = x$.<br\/> Ta c\u00f3 khai tri\u1ec3n ho\u00e0n ch\u1ec9nh l\u00e0: $x^2+40x+400=x^2+2.x.20+20^2$$=(x+20)^2$.<span>"}]}],"id_ques":428},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai3/lv3/img\/3.jpg' \/><\/center> So s\u00e1nh $A=\\left( 2+1 \\right)\\left( {{2}^{2}}+1 \\right)\\left( {{2}^{4}}+1 \\right)$$\\left( {{2}^{8}}+1 \\right)\\left( {{2}^{16}}+1 \\right)$ v\u1edbi $B={{2}^{32}}-1$ ","select":["A. A > B","B. A < B","C. A = B"],"hint":"Ta \u0111i khai tri\u1ec3n $A$: Nh\u00e2n $A$ v\u1edbi $(2-1)$ th\u00ec gi\u00e1 tr\u1ecb kh\u00f4ng \u0111\u1ed5i. <br\/> Khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $x^2-y^2=(x+y)(x-y)$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> Ta \u0111i khai tri\u1ec3n A b\u1eb1ng c\u00e1ch nh\u00e2n $A$ v\u1edbi $(2-1)$ th\u00ec gi\u00e1 tr\u1ecb kh\u00f4ng \u0111\u1ed5i. <br\/> <b> B\u01b0\u1edbc 2: <\/b> Khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $x^2-y^2=(x+y)(x-y)$.<br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh k\u1ebft qu\u1ea3 khai tri\u1ec3n \u0111\u00f3 v\u1edbi $B$.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/>$A=\\left( 2+1 \\right)\\left( {{2}^{2}}+1 \\right)\\left( {{2}^{4}}+1 \\right)\\left( {{2}^{8}}+1 \\right)\\left( {{2}^{16}}+1 \\right) $<br\/>$ =1.\\left( 2+1 \\right)\\left( {{2}^{2}}+1 \\right)\\left( {{2}^{4}}+1 \\right)\\left( {{2}^{8}}+1 \\right)\\left( {{2}^{16}}+1 \\right) $<br\/>$ =\\left( 2-1 \\right)\\left( 2+1 \\right)\\left( {{2}^{2}}+1 \\right)\\left( {{2}^{4}}+1 \\right)\\left( {{2}^{8}}+1 \\right)\\left( {{2}^{16}}+1 \\right) $<br\/>$ =\\left( {{2}^{2}}-1 \\right)\\left( {{2}^{2}}+1 \\right)\\left( {{2}^{4}}+1 \\right)\\left( {{2}^{8}}+1 \\right)\\left( {{2}^{16}}+1 \\right) $<br\/>$ =\\left( {{2}^{4}}-1 \\right)\\left( {{2}^{4}}+1 \\right)\\left( {{2}^{8}}+1 \\right)\\left( {{2}^{16}}+1 \\right) \\\\ = (2^8 - 1)(2^8 + 1)(2^{16} + 1) \\\\ = (2^{16} - 1)(2^{16} + 1) \\\\ ={{2}^{32}}-1 $<br\/>Do \u0111\u00f3 $A = B$. <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span>","column":3}]}],"id_ques":429},{"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/bai3/lv3/img\/4.jpg' \/><\/center> So s\u00e1nh $2005.2007$ v\u1edbi $2006^2$","select":["A. <","B. >","C. ="],"hint":"Ta \u0111i khai tri\u1ec3n $2005.2007$, t\u00e1ch: $2005=2006-1$, $2007=2006+1$. <br\/> Khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $x^2-y^2=(x+y)(x-y)$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ta \u0111i khai tri\u1ec3n $2005.2007$, t\u00e1ch: $2005=2006-1$, $2007=2006+1$. <br\/> <b> B\u01b0\u1edbc 2: <\/b> Khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $x^2-y^2=(x+y)(x-y)$.<br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh k\u1ebft qu\u1ea3 khai tri\u1ec3n \u0111\u00f3 v\u1edbi $2006^2$<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3: <br\/>$\\begin{align} 2005.2007& =\\left( 2006-1 \\right)\\left( 2006+1 \\right) \\\\ & ={{2006}^{2}}-{{1}^{2}} \\\\ & ={{2006}^{2}}-1<{{2006}^{2}} \\\\ \\end{align}$ <br\/> Do \u0111\u00f3: $2005.2007\\,\\,<\\,\\,{{2006}^{2}}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":3}]}],"id_ques":430}],"lesson":{"save":0,"level":3}}