Question

Помогите РЕШИТЬ 2 НОМЕРА
ДАЮ МНОГО БАЛОВ

Answer 1

Вроде бы так. Надеюсь правильно

Answer 2

$1)3\sqrt{5}=\sqrt{3^{2}*5}=\sqrt{45}\\\\5\sqrt{2}=\sqrt{5^{2}*2}=\sqrt{50}\\\\\sqrt{45}<\sqrt{50}\Rightarrow3\sqrt{5}<5\sqrt{2}$

$2)4\sqrt{\frac{3}{8}}=\sqrt{4^{2}*\frac{3}{8}}=\sqrt{16*\frac{3}{8}}=\sqrt{2*3}=\sqrt{6}\\\\\frac{1}{5}\sqrt{150}=\sqrt{(\frac{1}{5})^{2}*150}=\sqrt{\frac{1}{25}*150}=\sqrt{6} \\\\4\sqrt{\frac{3}{8}}=\frac{1}{5}\sqrt{150}$

$1)\frac{x-9}{\sqrt{x}+3 }=\frac{(\sqrt{x})^{2}-3^{2}}{\sqrt{x}+3}=\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}+3 }=\sqrt{x}-3\\\\2)\frac{5+2\sqrt{5}}{\sqrt{5}}=\frac{(\sqrt{5})^{2}+2\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}(\sqrt{5}+2)}{\sqrt{5}}=\sqrt{5}+2$

$3)\frac{a-1}{a-2\sqrt{a}+1}=\frac{(\sqrt{a})^{2}-1^{2}}{(\sqrt{a})^{2}-2\sqrt{a}+1}=\frac{(\sqrt{a}-1)(\sqrt{a}+1)}{(\sqrt{a}-1)^{2}}=\frac{\sqrt{a}+1}{\sqrt{a}-1}$