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Ta có 2012 chia cho 3 dư 2 => 2012^2 chia cko 3 dư 4
2013 chia cko 3 dư 0 => 2013^2 chia cko 3 dư 0
=> 2012^2x2013^2 chia cko 3 dư 0
=> A = 2012^2 + 2012^2x2013^2 + 2013^2 chia cko 3 dư 4 mak 4 chia 3 dư 1 suy ra A chia cko 3 dư 1
Mà số chính phương khi chia cko 3 chỉ có thể có số dư là 0 hoặc 1
Vậy A là số chính phương
Chỗ nào hong hiểu bn hỏi mk nhea =))
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\(a,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0.abc=0\)
Mà \(a+b+c=1=>\left(a+b+c\right)^2=1=>a^2+b^2+c^2+2ab+2bc+2ac=1\)
\(=>a^2+b^2+c^2+2\left(ab+bc+ac\right)=1=>a^2+b^2+c^2=1-0=1\) (vì ab+bc+ac=0)
\(b,S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3=\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3\)
\(=2014.\frac{1}{2014}-3=1-3=-2\)
Vậy.....................
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1/ Xét t/g ABC có DE // BC ; DE cắt AB,AC lần lượt lại D và E
=> \(\dfrac{AE}{AC}=\dfrac{AD}{AB}\) (đ.l Thales)
=> \(\dfrac{5}{8}=\dfrac{4}{4+DB}\)
=> 4 + DB = 6,4
=> DB = 2,4
2/ Chắc bạn viết nhầm gì đó?
3/ t/g AFG có HI // FG
=> \(\dfrac{HI}{FG}=\dfrac{AH}{AF}=\dfrac{AI}{AG}\) (Hệ quả đ/l Thales
=> \(\dfrac{AI}{AG}=\dfrac{3}{5,2}=\dfrac{2}{AF}\)
=> AF = \(\dfrac{10,4}{3}\)
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\(\text{a) Ta có:}xy=1\Rightarrow\hept{\begin{cases}2xy=2\\-2xy=-2\end{cases}}\)
\(\text{Ta lại có: }x^2+y^2=2\Rightarrow\hept{\begin{cases}x^2+y^2+2xy=2+2=4\\x^2+y^2-2xy=2-2=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=4\\\left(x-y\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x+y=\pm2\\x-y=0\end{cases}}}\)
\(\text{b) Ta có: }x+y=5\)
\(\Rightarrow\left(x+y\right)^2=25\)
\(\Rightarrow x^2+2xy+y^2=25\)
\(\Rightarrow x^2+4+y^2=25\)
\(\Rightarrow x^2+y^2=21\)
\(\text{b) Ta có: }x^2+y^2=21\)
\(\Rightarrow x^2-2xy+y^2=21-2xy\)
\(\Rightarrow\left(x-y\right)^2=21-4\)
\(\Rightarrow\left(x-y\right)^2=17\)
\(\Rightarrow x-y=\pm\sqrt{17}\)
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a. ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\times\left(-6\right)=13\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3\times\left(-6\right)\times1=19\)
\(x^5+y^5=\left(x+y\right)\left[x^4-x^3y+x^2y^2-xy^3+y^4\right]\)
\(=\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2-xy\left(x^2+y^2\right)\right]=1.\left(13^2-\left(-6\right)^2-\left(-6\right).13\right)=211\)
b.\(x^2+y^2=\left(x-y\right)^2+2xy=1+2\times6=13\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+6.3.1=19\)
\(x^5-y^5=\left(x-y\right)\left[\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\right]\)
\(=\left(x-y\right)\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=1.\left(13^2-6^2+6.13\right)=211\)
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1)
Ta có: x+y=2
nên \(\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow2xy=2\)
hay xy=1
Ta có: \(x^3+y^3\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=2^3-3\cdot1\cdot2\)
=2
2)\(x^2+y^2=\left(x+y\right)^2-2xy=8^2-2\cdot\left(-20\right)=104\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=8^3-3\cdot\left(-20\right)\cdot8=512+480=992\)
\(x^2+y^2+xy=\left(x+y\right)^2-xy=8^2-\left(-20\right)=64+20=84\)
\(x^{671}+y^{671}=1\Rightarrow\left(x^{671}+y^{671}\right)^2=x^{1342}+2.x^{671}.y^{671}+y^{1342}\)\(=1\)
Mà \(x^{1342}+y^{1342}=2\) \(\Rightarrow x^{671}.y^{671}=\dfrac{-1}{2}\)
Mặt khác: \(\left(x^{671}+y^{671}\right)^3=x^{2013}+3x^{671}y^{671}\left(x^{671}+y^{671}\right)+y^{2013}=1\)
Hay \(x^{2013}+y^{2013}-\dfrac{3}{2}.1=1\Rightarrow x^{2013}+y^{2013}=1+\dfrac{3}{2}=\dfrac{5}{2}\)