Đáp án+Giải thích các bước giải:
$\begin{array}{l}\begin{cases}x^2=\dfrac{y}{2}+\dfrac{1}{2y}\\y^2=\dfrac{x}{2}+\dfrac{1}{2x}\\\end{cases}\\↔x^2-y^2=\dfrac{y}{2}+\dfrac{1}{2y}-(\dfrac{x}{2}+\dfrac{1}{2x})\\↔(x-y)(x+y)=\dfrac{y-x}{2}+\dfrac{1}{2}(\dfrac{1}{y}-\dfrac{1}{x})\\↔(x-y)(x+y)+\dfrac{x-y}{2}=\dfrac{1}{2}.\dfrac{x-y}{xy}\\↔(x-y)(x+y+\dfrac{1}{2}-\dfrac{1}{2xy})=0\\↔\left[ \begin{array}{l}x=y\\x+y+\dfrac{1}{2}-\dfrac{1}{2xy}=0(1)\end{array} \right.\\(1)↔x+y=\dfrac{1}{2}(\dfrac{1}{xy}-1)\\↔x=\dfrac{1}{2}(\dfrac{1}{xy}-1)-y\\\text{vậy HPT có nghiệm}(x,y)=(x,x),(\dfrac{1}{2}(\dfrac{1}{xy}-1)-y,y)\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$
