1. If $x$ and $y$ are real numbers such that $\frac{x+22}{y}+\frac{290}{x y}=\frac{26-y}{x}$, find $x y$.
(1 mark)若實數 $x 、 y$ 滿足 $\frac{x+22}{y}+\frac{290}{x y}=\frac{26-y}{x}$, 求 $x y$ 。
1. -143
1. Multiplying both sides by $x y$, we get $x^{2}+22 x+290=26 y-y^{2}$. Since $290=11^{2}+13^{2}$, this is equivalent to $(x+11)^{2}+(y-13)^{2}=0$. As $x$ and $y$ are real, we must have $x=-11$ and $y=13$, giving $x y=-143$.