ПОМОГИТЕ ПЖ,напишите решение на листе.#245

Решите задачу:

$\frac{a-2b}{5b+3a} =5\; \; \; \Rightarrow \\\\\frac{4b-2a}{5b+3a} = \frac{-2(a-2b)}{5b+3a} =-2\cdot 5=-10\\\\\\ \frac{5a-10b}{15b+9a} = \frac{5(a-2b)}{3(5b+3a)} = \frac{5}{3}\cdot 5=\frac{25}{3}\\\\\\ \frac{a^2-4ab+4b^2}{9a^2+30ab+25b^2} =\frac{(a-2b)^2}{(3a+5b)^2} =(\frac{a-2b}{5b+3a})^2=5^2=25$

$\frac{2a^2b^2-a^3b}{6a^3b+10a^2b^2} =\frac{a^2b(2b-a)}{2a^2b(3a+5b)}=\frac{-(a-2b)}{2(5b+3a)}=-\frac{1}{2}\cdot 5=-2,5\\\\\\ \frac{15b^4+9ab^3}{ab^3-2b^4} =\frac{3b^3(5b+3a)}{b^3(a-2b)}=\frac{3}{\frac{a-2b}{5b+3a}} =\frac{3}{5}=0,6\\\\\\ \Big (\frac{5b+3a}{2b-a}\Big )^3=\Big (\frac{-1}{\frac{a-2b}{5b+3a}}\Big )^3 =\frac{-1}{5^3}=-\frac{1}{125}=-0,008$

ПОМОГИТЕ ПЖ,напишите решение ** листе.#245