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\(\frac{xy}{x^2+y^2}=\frac{5}{8}\)
\(\Rightarrow5\left(x^2+y^2\right)=8xy\)
Ta có : \(P=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}=\frac{5\left(x^2+y^2-2xy\right)}{5\left(x^2+y^2+2xy\right)}\)
\(=\frac{5\left(x^2+y^2\right)-10xy}{5\left(x^2+y^2\right)+10xy}=\frac{8xy-10xy}{8xy+10xy}=\frac{-2xy}{18xy}=\frac{-1}{9}\)
Ta có: \(P=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}=\frac{\frac{x^2+y^2-2xy}{x^2+y^2}}{\frac{x^2+y^2+2xy}{x^2+y^2}}=\frac{\frac{x^2+y^2}{x^2+y^2}-\frac{2xy}{x^2+y^2}}{\frac{x^2+y^2}{x^2+y^2}+\frac{2xy}{x^2+y^2}}\)
\(=\frac{1-\frac{2xy}{x^2+y^2}}{1+\frac{2xy}{x^2+y^2}}=\frac{1-\frac{2.5}{8}}{1+\frac{2.5}{8}}=\frac{-1}{9}\)
Vậy \(P=\frac{-1}{9}\)
\(\frac{x^4-xy^3}{2xy+y^2}:\frac{x^3+x^2y+xy^2}{2x+y}\)
\(=\frac{x\left(x^3-y^3\right)}{y\left(2x+y\right)}.\frac{2x+y}{x^3+x^2y+xy^2}\)
\(=\frac{x\left(x-y\right)\left(x^2+xy+y^2\right)\left(2x+y\right)}{xy\left(2x+y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{x-y}{y}\)
\(\frac{x^4-xy^3}{2xy+y^2}:\frac{x^3+x^2y+xy^2}{2x+y}\)
\(=\frac{x\left(x^3-y^3\right)}{y\left(2x+y\right)}:\frac{x\left(x^2+xy+y^2\right)}{2x+y}\)
\(=\frac{x\left(x-y\right)\left(x^2+xy+y^2\right)}{y\left(2x+y\right)}:\frac{x\left(x^2+xy+y^2\right)}{2x+y}\)
\(=\frac{x\left(x-y\right)\left(x^2+xy+y^2\right)}{y\left(2x+y\right)}.\frac{2x+y}{x\left(x^2+xy+y^2\right)}\)
\(=\frac{x-y}{y}\)
\(\frac{xy}{x^2+y^2}=\frac{3}{8}\Rightarrow xy=\frac{3}{8}\left(x^2+y^2\right)\)
=>\(A=\frac{x^2+y^2+\frac{3}{4}\left(x^2+y^2\right)}{x^2+y^2-\frac{3}{4}\left(x^2+y^2\right)}=\frac{\frac{7}{4}\left(x^2+y^2\right)}{\frac{1}{4}\left(x^2+y^2\right)}=7\)
