Question

[Ngọc Minh]

a) cho \(\dfrac{xy}{x^2+y^2}=\dfrac{5}{8}\) . Tính \(A=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}\)

b) cho \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) . Tính \(B=\dfrac{x^2+y^2+z^2}{\left(ã+by+cz\right)^2}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

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

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

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