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Вот эти 2 задания

Answer 1

В1.

$\frac{\sqrt{6}-2-\sqrt{10}  }{\sqrt{3}-\sqrt{2}-\sqrt{5} } = \frac{(\sqrt{6} -2-\sqrt{10})*(\sqrt{3}-\sqrt{2} + \sqrt{5} ) }{-2\sqrt{6} } = \frac{\sqrt{18}-\sqrt{12} +\sqrt{30}  -2\sqrt{3} + 2\sqrt{2} - 2\sqrt{5} -\sqrt{30}+\sqrt{20} -\sqrt{50}  }{-2\sqrt{6} } =$ $\frac{3\sqrt{2}-2\sqrt{3} -2\sqrt{3} + 2\sqrt{2} - 2\sqrt{5}+2\sqrt{5}   -\sqrt{50} }{-2\sqrt{6} } = \frac{3\sqrt{2}-2\sqrt{3} -2\sqrt{3} + 2\sqrt{2}    -5\sqrt{2} }{-2\sqrt{6} } = \frac{0-4\sqrt{3} }{-2\sqrt{6} } = \frac{-4\sqrt{3} }{-2\sqrt{6} } = \frac{2}{\sqrt{2} } = \sqrt{2}$

С1.

$\frac{\sqrt{y} }{\sqrt{x} -2\sqrt{y} } = \frac{\sqrt{y}* (\sqrt{x} + 2\sqrt{y} ) }{(\sqrt{x}-2\sqrt{y} ) * (\sqrt{x} + 2\sqrt{y} ) } =\frac{\sqrt{xy} +2y}{x-4y}$