Question

cho x>y>0 và x^2 +y^2/xy = 5/2. Tính giá trị biểu thức E= 3x + 2y/2x-3y

Answer 1

\(\dfrac{x^2+y^2}{xy}=\dfrac{5}{2}\Leftrightarrow2x^2+2y^2-5xy=0\)

\(\Leftrightarrow2x^2+2y^2-4xy-xy=0\)

\(\Leftrightarrow\left(2x^2-xy\right)-\left(4xy-2y^2\right)=0\)

\(\Leftrightarrow x\left(2x-y\right)-2y\left(2x-y\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(2x-y\right)=0\)

Ta có: \(x>y>0\Leftrightarrow x+x>y+0\Leftrightarrow2x>y\Leftrightarrow2x-y>0\)

Vậy \(x-2y=0\Leftrightarrow x=2y\)

\(E=\dfrac{3x+2y}{2x-3y}=\dfrac{6y+2y}{4y-3y}=\dfrac{8y}{y}=8\)