Question

Cho \(x^2+y^2=\dfrac{50}{7}xy\) với y>x>0 . Giá trị của biểu thức \(P=\dfrac{x-y}{x+y}\) .. là

Answer 1

\(P=\dfrac{x-y}{x+y}\)

=> \(P^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{\left(x-y\right)^2}{\left(x+y\right)^2}=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}\) (*)

Thay x² + y² = \(\dfrac{50}{7}xy\) vào (*), ta có:

\(P^2=\dfrac{\dfrac{50}{7}xy-2xy}{\dfrac{50}{7}xy+2xy}=\dfrac{\dfrac{36}{7}xy}{\dfrac{64}{7}xy}=\dfrac{9}{16}\)

=> \(P=\sqrt{\dfrac{9}{16}}=\sqrt{\left(\pm\dfrac{3}{4}\right)^2}=\pm\dfrac{3}{4}\)

mà y > x > 0

=> P = 0,75