Question

Помогите пожалуйста номер 10
Zombynella помогите пожалуйста

Answer 1

Ответ:

В решении.

Объяснение:

На фото.

Answer 2

Ответ:

$9)\ \ \displaystyle \frac{5+\sqrt2}{\sqrt3}=\frac{(5+\sqrt2)\sqrt3}{\sqrt3\cdot \sqrt3}=\frac{(5+\sqrt2)\sqrt3}{3}=\frac{5\sqrt3+\sqrt6}{3}\\\\\\\frac{6+\sqrt3}{\sqrt6}=\frac{\sqrt6(6+\sqrt3)}{\sqrt6\cdot \sqrt6}=\frac{\sqrt6(6+\sqrt3)}{6}=\frac{6\sqrt6+\sqrt{18}}{6}=\frac{6\sqrt6+3\sqrt2}{6}=\frac{2\sqrt6+\sqrt2}{2}\\\\\\\frac{5+2\sqrt{12}}{2\sqrt{12}}=\frac{5+2\sqrt{4\cdot 3}}{2\sqrt{4\cdot 3}}=\frac{5+4\sqrt3}{4\sqrt3}=\frac{\sqrt3\, (5+4\sqrt3)}{12}$

$\displaystyle \frac{12-4\sqrt3}{6\sqrt3}=\frac{\sqrt3\, (12-4\sqrt3)}{18}\\\\\\\frac{\sqrt5-2}{5-\sqrt{10}}=\frac{(\sqrt5-2)(5+\sqrt10)}{(5-\sqrt{10})(5+\sqrt{10})}=\frac{(\sqrt5-2)(5+\sqrt{10})}{25-10}=\frac{(\sqrt5-2)(5+\sqrt{10})}{15}\\\\\\\frac{2\sqrt3+1}{7-\sqrt3}=\frac{(2\sqrt3+1)(7+\sqrt3)}{49-3}=\frac{(2\sqrt3+1)(7+\sqrt3)}{46}$

$\displaystyle \frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{a}+\sqrt{b}}{a-b}\\\\\\\frac{3\sqrt{x}}{1+\sqrt{x}}=\frac{3\sqrt{x}\, (1-\sqrt{x})}{(1+\sqrt{x})(1-\sqrt{x})}=\frac{3\sqrt{x}\, (1-\sqrt{x})}{1-x}$

$10)\ \ (\sqrt{a}+\sqrt{b})^2=a+2\sqrt{ab}+b\\\\(\sqrt3+2)^2=3+4\sqrt3+4=7+4\sqrt3\\\\(\sqrt5-\sqrt2)(\sqrt5+\sqrt2)=5-2=3\\\\(2\sqrt2+\sqrt{20})(\sqrt8-2\sqrt5)=(\sqrt8+\sqrt{20})(\sqrt8-\sqrt{20})=8-20=-12\\\\\sqrt5\, (\sqrt{10}+\sqrt{125}+\sqrt{16,2})=\sqrt{50}+\sqrt{625}+\sqrt{81}=5\sqrt2+25+9=5\sqrt2+34\\\\\\(\sqrt3+\sqrt{x})^3=(\sqrt3)^3+3\, (\sqrt3)^2\, x+3\sqrt3\, (\sqrt{x})^2+(\sqrt{x})^3=\\\\=3\sqrt3+9x+3x\sqrt3+x\sqrt{x}$