a.

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

= x^2 + y^2 - 2xy + x^2 + y^2 + 2xy

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

= 2x^2 + 2y^2

b.

(x^2 + y^2 - 2xy) - (x^2 + y^2 + 2xy)

= x^2 + y^2 - 2xy - x^2 - y^2 - 2xy

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

= -4xy

\(\dfrac{x}{x^2+2xy+y^2}+\dfrac{2y}{x+y}+\dfrac{y}{x^2+2xy+y^2}\)

\(=\dfrac{x+y}{\left(x+y\right)^2}+\dfrac{2y}{x+y}\)

\(=\dfrac{1}{x+y}+\dfrac{2y}{x+y}=\dfrac{2y+1}{x+y}\)

Ta có:

 VT=(x2+y2)2−(2xy)2VT=(x2+y2)2−(2xy)2

=(x2+y2−2xy)(x2+y2+2xy)=(x2+y2−2xy)(x2+y2+2xy)

=(x−y)2(x+y)2=VP=(x−y)2(x+y)2=VP

⇒đpcm⇒đpcm

Bexiu bị j ấy

a: \(\dfrac{xy}{x^2+y^2}=\dfrac{5}{8}\)

=>\(\dfrac{xy}{5}=\dfrac{x^2+y^2}{8}=k\)

=>\(xy=5k;x^2+y^2=8k\)

\(A=\dfrac{8k-2\cdot5k}{8k+2\cdot5k}=\dfrac{-2}{18}=\dfrac{-1}{9}\)

b: Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\)

=>x=a*k; y=b*k; z=c*k

\(B=\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{\left(a\cdot ak+b\cdot bk+c\cdot ck\right)^2}\)

\(=\dfrac{k^2\cdot\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)

kinh nhờ học nhà thầy Khánh à ?

mấy bạn biết thầy Khánh ak thầy mk đó

P=3x^2y^2-x^3-2xy+6y^2+3x^2+2xy-6y^2

=3x^2y^2+3x^2-x^3

=3*(-2)^2*(-2)^2+3*(-2)^2-(-2)^3

=68