Помогите пожалуйста) № 1,2,5

Question 1

Помогите пожалуйста)
№ 1,2,5

Question 2

1) а) $\sqrt{0,0025}= \sqrt{25*0,0001} =5*0,01 = 0,05$
б) $\sqrt{256*9*36}=16*3*6=288$
в) $\sqrt{\frac{25*324}{529*49} }= \frac{5*18}{23*7}= \frac{90}{161}$
г) $\sqrt{( \sqrt{3}-2 )^2}= | \sqrt{3} -2|=2- \sqrt{3}$
$2- \sqrt{3}=2-1,732=0,268; 0,26\ \textless \ 2- \sqrt{3}\ \textless \ 0,27$
д) $\sqrt[3]{( \sqrt{3}-2 )^3} = \sqrt{3}-2$
$-0,27\ \textless \ \sqrt{3}-2\ \textless \ -0,26$

2) а) $\sqrt[4]{32a^4b^3}=2a \sqrt[4]{2b^3}$
б) $\sqrt{25a^2b^3}=-5ab \sqrt{b}$
в) $\sqrt{(x-3)^2+12x}= \sqrt{x^2-6x+9+12x}= \sqrt{x^2+6x+9}=$
$= \sqrt{(x+3)^2} =|x+3|$
г) $\sqrt[3]{x^3y^6}=xy^2$
д) $\sqrt{169x^3y^2} =-13xy \sqrt{x}$
е) $\sqrt[4]{8a^5b^6}=-ab \sqrt[4]{8ab^2}$

5) а) $\frac{1}{ \sqrt[3]{x^2y^3} } = \frac{1}{y \sqrt[3]{x^2} } = \frac{ \sqrt[3]{x} }{xy}$
б) $\frac{1}{2 \sqrt{5}- \sqrt{7}}= \frac{2 \sqrt{5}+ \sqrt{7}}{(2 \sqrt{5}- \sqrt{7})(2 \sqrt{5}+ \sqrt{7})}= \frac{2 \sqrt{5}+ \sqrt{7}}{4*5-7}= \frac{2 \sqrt{5}+ \sqrt{7}}{13}$
в) $\frac{1}{ \sqrt[3]{5} - \sqrt[3]{2} } = \frac{ \sqrt[3]{5^2}+ \sqrt[3]{5*2}+ \sqrt[3]{2^2}}{(\sqrt[3]{5} - \sqrt[3]{2})(\sqrt[3]{5^2}+ \sqrt[3]{5*2}+ \sqrt[3]{2^2})} = \frac{\sqrt[3]{25}+ \sqrt[3]{10}+ \sqrt[3]{4}}{5-2}= \frac{\sqrt[3]{25}+ \sqrt[3]{10}+ \sqrt[3]{4}}{3}$
г) $\frac{5}{ \sqrt{13}- \sqrt{18}}= \frac{5(\sqrt{13}+ \sqrt{18})}{(\sqrt{13}- \sqrt{18})(\sqrt{13}+ \sqrt{18})} = \frac{5(\sqrt{13}+ \sqrt{18})}{13-18}=-(\sqrt{13}+ \sqrt{18})$
д) $\frac{1}{1+ \sqrt{2}+ \sqrt{7}} = \frac{1+ \sqrt{2}- \sqrt{7}}{(1+ \sqrt{2}+ \sqrt{7})(1+ \sqrt{2}- \sqrt{7})} = \frac{1+ \sqrt{2}- \sqrt{7}}{(1+ \sqrt{2})^2-7}= \frac{1+ \sqrt{2}- \sqrt{7}}{1+2 \sqrt{2}+2-7 }=$
$= \frac{1+ \sqrt{2}- \sqrt{7}}{2 \sqrt{2}-4}= \frac{(1+ \sqrt{2}- \sqrt{7})(2 \sqrt{2}+4)}{4*2-16}= \frac{(1+ \sqrt{2}- \sqrt{7})(2 \sqrt{2}+4)}{-8}=$
$=-\frac{(1+ \sqrt{2}- \sqrt{7})(\sqrt{2}+2)}{4}= \frac{(\sqrt{7}-\sqrt{2}-1)(\sqrt{2}+2)}{4}$