Do \(\left(a+b\right)^2=a^2+2ab+b^2\)nên \(a^2+b^2=\left(a+b\right)^2-2ab=9^2-2.20=41\)

Ta có: \(\left(a-b\right)^2=a^2-2ab+b^2=41-2.20=1\Rightarrow a-b=1\)hoặc \(a-b=-1\)

Với \(a-b=1\) thì \(\left(a-b\right)^{2015}=1^{2015}=1\) 

Với \(a-b=-1\) thì \(\left(a-b\right)^{2015}=\left(-1\right)^{2015}=-1\)

ai giup mik dc ko ak pls mik can gap

\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)

ọi k là một số nguyên, theo đề ta có: 
a=3k+1 
b=3k+2 
ab=(3k+1)(3k+2)=9k^2+9k+2 
vì 9k^2 và 9k chia hết cho 3 
nên ab chia 3 dư 2

- Vì a chia cho 3 dư 1 nên a = 3m + 1 ( m \(\in\)N )

- Vì b chia cho 3 dư 2 nên b = 3n + 2 ( n\(\in\)N )

Ta có :

a . b = ( 3m + 1 ) ( 3n + 2 )

        = 3m . 3n + 3m . 2 + 1 . 3n + 1 . 2

        = ( 9 mn + 6m + 3n ) + 2

        = 3 ( 3mn + 2m + n ) + 2 ....

Vậy ab chia cho 3 dư 2 .

\(\left(a-b\right)^2=a^2-2ab+b^2=a^2+2ab+b^2-4ab=\left(a+b\right)^2-4ab\)

= 5²-4.2=25-8=17

a) Ta có:

\(VT=\left(a-b\right)^2\)

\(=a^2-2ab+b^2\)

\(=a^2+2ab+b^2-4ab\)

\(=\left(a+b\right)^2-4ab=VP\left(dpcm\right)\)

b) Ta có:

\(VT=\left(x+y\right)^2+\left(x-y\right)^2\)

\(=x^2+2xy+y^2+x^2-2xy+y^2\)

\(=\left(x^2+y^2\right)+\left(x^2+y^2\right)\)

\(=2\left(x^2+y^2\right)=VP\left(dpcm\right)\)

a) ta có : \(\left(a+b\right)^2=a^2+2ab+b^2=a^2-2ab+b^2+4ab\)

\(=\left(a-b\right)^2+4ab\left(đpcm\right)\)

b) ta có : \(a+b=9\Rightarrow\left(a+b\right)^2=9^2=81\)

\(\Leftrightarrow a^2+2ab+b^2=81\Leftrightarrow a^2-2ab+b^2+4ab=81\)

\(\Leftrightarrow\left(a-b\right)^2+4ab=81\) ............................................(1)

thay \(ab=20\) vào (1)

ta có (1) \(\Leftrightarrow\left(a-b\right)^2+4\left(20\right)=81\)

\(\Leftrightarrow\left(a-b\right)^2+80=81\Leftrightarrow\left(a-b\right)^2=81-80=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=1\\a-b=-1\end{matrix}\right.\)

th1: \(a-b=1\Rightarrow\left(a-b\right)^{2015}=\left(1\right)^{2015}=1\)

th2: \(a-b=-1\Rightarrow\left(a-b\right)^{2015}=\left(-1\right)^{2015}=-1\)

vậy \(\left(a-b\right)^{2005}=\pm1\)