Question

[Lê Tài Bảo Châu]

\(P=\frac{x-y}{x+y}\)\(\Rightarrow P^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}\)

Ta có: \(2x^2+2y^2=5xy\left(1\right)\)

\(\Leftrightarrow2x^2-4xy+2y^2=xy\)

\(\Leftrightarrow2\left(x-y\right)^2=xy\)

\(\Leftrightarrow\left(x-y\right)^2=\frac{xy}{2}\)(2)

Từ \(\left(1\right)\Rightarrow2x^2+4xy+2y^2=9xy\)

\(\Leftrightarrow2\left(x+y\right)^2=9xy\)

\(\Leftrightarrow\left(x+y\right)^2=\frac{9xy}{2}\)(3)

Thay (2)và (3) vào \(P^2\)ta được:

\(P^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{xy}{2}.\frac{2}{9xy}=\frac{1}{9}\)

Nhận thấy \(y>x>0\Rightarrow x-y< 0\Rightarrow\frac{x-y}{x+y}< 0\Rightarrow P< 0\)

\(\Rightarrow P=\frac{-1}{3}\)

Answer 1

[๖ۣۜRᶤℵ﹏❖(๖ۣۜBảo)]

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