Question

Срочно помогите даю 100 баллов​

Answer 1

Ответ:

$a)\ \ \displaystyle \frac{\sqrt2+\sqrt8-\sqrt{32}}{-\sqrt2}=\frac{\sqrt2+2\sqrt2-4\sqrt2}{-\sqrt2}=\frac{-\sqrt2}{-\sqrt2}=1\\\\\\b)\ \ \frac{\sqrt3-\sqrt{27}+\sqrt{48}}{\sqrt3}=\frac{\sqrt3-3\sqrt3+4\sqrt3}{\sqrt3}=\frac{2\sqrt3}{\sqrt3}=2\\\\\\c)\ \ \frac{10-\sqrt{125}}{\sqrt{5(2-\sqrt5)^2}}=\frac{10-5\sqrt5}{\sqrt5\cdot |2-\sqrt5|}=\frac{-5(\sqrt5-2)}{\sqrt5(\sqrt5-2)}=\frac{-\sqrt5}{1}=-\sqrt5$

$\displaystyle d)\ \ \frac{(\sqrt2+1)^3+(\sqrt2-1)^3}{\sqrt8}=\frac{2\sqrt2+6+3\sqrt2+1+2\sqrt2-6+3\sqrt2-1}{\sqrt8}=\\\\\\=\frac{4\sqrt2+6\sqrt2}{\sqrt8}=\frac{10\sqrt2}{2\sqrt2}=\frac{10}{2}=5$