Question

Cho \(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\) với điều kiện 0 < x < y

Tính giá trị biểu thức \(B=\dfrac{x-y}{x+y}\)

Answer 1

\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\Leftrightarrow x^2+y^2=\dfrac{10}{3}xy\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+2xy=\dfrac{10}{3}xy+2xy\\x^2+y^2-2xy=\dfrac{10}{3}xy-2xy\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=\dfrac{16}{3}xy\\\left(x-y\right)^2=\dfrac{4}{3}xy\end{matrix}\right.\)

Do \(0< x< y\Rightarrow\left\{{}\begin{matrix}x-y>0\\x+y>0\end{matrix}\right.\) \(\Rightarrow B>0\)

\(B^2=\dfrac{\left(x-y\right)^2}{\left(x+y\right)^2}=\dfrac{\dfrac{4}{3}xy}{\dfrac{16}{3}xy}=\dfrac{1}{4}\Rightarrow B=\dfrac{1}{2}\)