GT=>(2x-y)(x-2y)=0
Do 0<x<y nên x-2y<0
Do đó 2x-y=0 hay 2x=y
Thay y=2x vào E đượcE=-3
Ta có: \(2\left(x^2+y^2\right)=5xy\)
\(x^2+y^2=\frac{5}{2}xy\)
\(E^2=\left(\frac{x+y}{x-y}\right)^2=\frac{\left(x+y\right)^2}{\left(x-y\right)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}\)
Hay: \(\frac{\frac{5}{2}xy+2xy}{\frac{5}{2}xy+2xy}=\frac{4,5xy}{0,5xy}=9\)
\(\Rightarrow E=\sqrt{9}=\pm3\)
vì 0<x<y
=>E=3
Ta có:\(2x^2+2y^2=5xy\Leftrightarrow2x^2+2y^2+4xy=9xy\Leftrightarrow2\left(x+y\right)^2=9xy\Leftrightarrow\left(x+y\right)^2=\frac{9xy}{2}\) (1)
Mặt khác \(2x^2+2y^2=5xy\Leftrightarrow2x^2+2y^2-4xy=xy\Leftrightarrow2\left(x-y\right)^2=xy\Leftrightarrow\left(x-y\right)^2=\frac{xy}{2}\) (2)
Từ (1) và (2) => \(\frac{\left(x+y\right)^2}{\left(x-y\right)^2}=\frac{\frac{9xy}{2}}{\frac{xy}{2}}\Leftrightarrow\left(\frac{x+y}{x-y}\right)^2=9\Leftrightarrow\frac{x+y}{x-y}=\pm3\)
Mà \(0< x< y\Rightarrow E=\frac{x+y}{x-y}=-3\)
Vậy E=-3
á mk nhầm nha bn E=-3 ms đúng,sorry nhiều!!!
