1. Calculate the value of the expression $(x+y)^{2}$, if $\frac{1}{x}-\frac{1}{y}=\frac{1}{2}$ and $y-x=1$.
Solution. From $1=\frac{2}{x}-\frac{2}{y}=2 \frac{y-x}{x y}=2 \frac{1}{x y}$ it follows that $x y=2$. Now,

$$
(x+y)^{2}=x^{2}+2 x y+y^{2}=x^{2}-2 x y+y^{2}+4 x y=(x-y)^{2}+4 x y=1^{2}+4 \cdot 2=9
$$