С 1 по 4 номер. Выбираю лучший ответ.

Question 1

Question 2

Решите задачу:

$1)\ \sqrt[3]{ab}=\sqrt[3]a\cdot\sqrt[3]b,\\a\ \&\ b\in \mathbb R\ (B)$

$2)\ 3\sqrt7=\sqrt{3^2\cdot7}=\sqrt{63},\\7\sqrt{3}=\sqrt{7^2\cdot3}=\sqrt{147},\\\sqrt{63}\ \textless \ \sqrt{147}\ \Longrightarrow\ 3\sqrt{7}\ \textless \ 7\sqrt{3};$

$3)\ \left(\sqrt8-3\sqrt2+\sqrt{10}\right)\left(\sqrt2+\sqrt{1,6}+3\sqrt{0,4}\right)=\\=\left(\sqrt{2^3}-3\sqrt2+\sqrt{10\right)\left(\sqrt2+\sqrt{4\cdot0,4}+3\sqrt{0,4}\right)=\\=\left(2\sqrt{2}-3\sqrt2+\sqrt{10}\right)\left(\sqrt2+2\sqrt{0,4}+3\sqrt{0,4}\right)=\\=\left(-\sqrt2+\sqrt{10}\right)\left(\sqrt2+5\sqrt{0,4}\right)=\left(-\sqrt2+\sqrt{10}\right)\left(\sqrt2+\sqrt{5^2\cdot0,4}\right)=\\=\left(-\sqrt2+\sqrt{10}\right)\left(\sqrt2+\sqrt{10}\right)=$
$=\left(\sqrt{10}-\sqrt2\right)\left(\sqrt{10}+\sqrt2\right)=\left(\sqrt{10}\right)^2-\left(\sqrt2\right)^2=10-2=8.$

$4)\ \frac{\sqrt x+1}{x\sqrt x+x+\sqrt x}:\frac{1}{x^2-\sqrt x}=\frac{\left(\sqrt x+1\right)\left(x^2-\sqrt x\right)}{x\sqrt x+x+\sqrt x}=\frac{x^2\sqrt x-x+x^2-\sqrt x}{x\sqrt x+x+\sqrt x}=\\\\=\frac{\left(x^2\sqrt x -\sqrt x\right)+\left(x^2-x\right)}{x\sqrt x+x+\sqrt x}=\frac{\sqrt x\left(x^2 -1)+x\left(x-1\right)}{x\sqrt x+x+\sqrt x}=\\\\=\frac{\sqrt x\left(x-1)(x+1)+x\left(x-1\right)}{x\sqrt x+x+\sqrt x}=\frac{(x-1)\left(\sqrt x(x+1)+x\right)}{x\sqrt x+x+\sqrt x}=$

$=\frac{(x-1)\left(x\sqrt x+\sqrt x+x\right)}{x\sqrt x+x+\sqrt x}=\frac{(x-1)\left(x\sqrt x+x+\sqrt x\right)}{x\sqrt x+x+\sqrt x}=x-1.$