Đáp án: $A=\dfrac{9\pm\sqrt{17}}{1\pm\sqrt{17}}$
Giải thích các bước giải:
Ta có:
$2x^2+y^2=5xy$
$\to 2x^2-5xy+y^2=0$
$\to x^2-\dfrac52xy+\dfrac12y^2=0$
$\to x^2-2x\cdot \dfrac54y+(\dfrac54y)^2=\dfrac{17}{16}y^2$
$\to (x-\dfrac54y)^2=\dfrac{17}{16}y^2$
$\to x-\dfrac54y=\pm\dfrac{y\sqrt{17}}{4}$
$\to x=\dfrac{5y\pm y\sqrt{17}}{4}$
$\to x=\dfrac{5\pm\sqrt{17}}{4}\cdot y$
$\to x+y=\dfrac{9\pm\sqrt{17}}{4}\cdot y$
$x-y=\dfrac{1\pm\sqrt{17}}{4}\cdot y$
$\to A=\dfrac{\dfrac{9\pm\sqrt{17}}{4}\cdot y}{\dfrac{1\pm\sqrt{17}}{4}\cdot y}$
$\to A=\dfrac{9\pm\sqrt{17}}{1\pm\sqrt{17}}$
