Question

помогите пожалуйста решить​

Answer 1

Ответ:

1. $4\sqrt{b}$

2.

а) $18$

б) $4a^2-b$

в) $5$

3. $\frac{1}{\sqrt{x}-\sqrt{3}}$

Пошаговое объяснение:

1. $\sqrt{36b}-\sqrt{16b}+2\sqrt{b}=$

$\sqrt{36}\sqrt{b}-\sqrt{16}\sqrt{b}+2\sqrt{b}=$

$6\sqrt{b}-4\sqrt{b}+2\sqrt{b}=$

$4\sqrt{b}$

2.

а) $(3\sqrt{8}+\sqrt{18})*\sqrt{2}=$

$(3\sqrt{4*2}+\sqrt{9*2})*\sqrt{2}=$

$(6\sqrt{2}+3\sqrt{2})*\sqrt{2}=$

$3(2\sqrt{2}+\sqrt{2})*\sqrt{2}=$

$3(3\sqrt{2})*\sqrt{2}=$

$9\sqrt{2}*\sqrt{2}=$

$9\sqrt{2*2}=$

$9*2=$

$18$

б) $(2a-\sqrt{b})(2a+\sqrt{b})=$

$(2a)^2-(\sqrt{b})^2=$

$4a^2-b$

в) $(\sqrt{3}+\sqrt{2})^2-\sqrt{24}=$

$((\sqrt{3})^2+2\sqrt{3}\sqrt{2}+(\sqrt{2})^2)-\sqrt{24}=$

$5+2\sqrt{3*2}-\sqrt{4*6}=$

$5+2\sqrt{6}-2\sqrt{6}=$

$5$

3. $\frac{\sqrt{x}+\sqrt{3}}{x-3}=$

$\frac{(\sqrt{x}+\sqrt{3})*(\sqrt{x}-\sqrt{3})}{(x-3)*(\sqrt{x}-\sqrt{3})}=$

$\frac{(\sqrt{x})^2-(\sqrt{3})^2}{(x-3)*(\sqrt{x}-\sqrt{3})}=$

$\frac{(x-3)}{(x-3)*(\sqrt{x}-\sqrt{3})}=$

$\frac{(x-3)*1}{(x-3)*(\sqrt{x}-\sqrt{3})}=$

$\frac{1}{\sqrt{x}-\sqrt{3}}$