Question

($\frac{2y}{x} - \frac{x}{2y}$):(2y-x) ,если x= $\frac{1}{9}$ y= $\frac{1}{6}$

Answer 1

$(\frac{2y}{x}-\frac{x}{2y}):(2y-x)=\frac{4y^2-x^2}{2xy}\cdot \frac{1}{2y-x}=\frac{(2y-x)(2y+x)}{2xy(2y-x)}=\\\\=\frac{2y+x}{2xy}=\frac{\frac{2}{6}+\frac{1}{9}}{\frac{2}{6\cdot 9}}=\frac{\frac{4}{9}}{\frac{1}{27}}=12$