Сократить дробь: x^2-20/x√5-10

Решите задачу:

$\frac{x^2-20}{x\sqrt5-10}=\frac{(x-\sqrt{20})(x+\sqrt{20})}{x\sqrt5-5\cdot 2}=\frac{(x-2\sqrt5)(x+2\sqrt5)}{\sqrt5(x-2\sqrt5)} = \frac{x+2\sqrt5}{\sqrt5} = \frac{x}{\sqrt5}+2=\\\\= \frac{x\sqrt5}{5}+2\\\\\\P.S.\; \; \sqrt{20} =\sqrt{4\cdot 5}= \sqrt{4}\cdot \sqrt{5}=2\sqrt{5}\\\\x\sqrt5-5\cdot 2=x\sqrt5-(\sqrt5)^2\cdot 2=x\sqrt5-\sqrt5\cdot \sqrt5\cdot 2=\sqrt5(x-2\sqrt5)$