Сократите дробь:1).2).3).

Решите задачу:

$\frac{a -\sqrt{3a} +3}{a \sqrt{a}+3 \sqrt{3} } = \frac{a -\sqrt{3a} +3}{ (\sqrt{a}+ \sqrt{3} )(a -\sqrt{3a} +3)} = \frac{1}{ \sqrt{a}+ \sqrt{3}} \\\ \frac{2 \sqrt{10}-5 }{4- \sqrt{10} } = \frac{\sqrt{10}(2 -0,5\sqrt{10}) }{\sqrt{10}(0,4\sqrt{10}-1)} = \frac{2 -0,5\sqrt{10} }{0,4\sqrt{10}-1} \\\ \frac{9-2 \sqrt{3} }{3 \sqrt{6}-2 \sqrt{2} } = \frac{ \sqrt{3}(3 \sqrt{3}-2) }{ \sqrt{2} (3 \sqrt{3} -2) } = \frac{ \sqrt{3}}{ \sqrt{2} } = \sqrt{1,5}$