Помогите пожалуйста с алгеброй!!

Question 1

Question 2

4) $\frac{3}{a}$ + 5 - $\frac{2}{a-b}$ = $\frac{3(a-b)+5a(a-b)-2a}{a(a-b)}$ = $\frac{3a-3b+5a^{2}-5ab-2a}{a(a-b)}$ = $\frac{a-3b+5a^{2}-5ab}{a(a-b)}$

6) $\frac{3a}{6a+8}$ - $\frac{1}{2}$ - $\frac{2}{4-3a}$ = $\frac{3a}{2(3a+4)}$ - $\frac{1}{2}$ + $\frac{2}{3a-4}$ = $\frac{3a(3a-4)-(9a^{2}-16)+2(3a+4)}{2(3a+4)(3a-4)}$ = $\frac{9a^{2}-12a-9a^{2}+16+6a+8}{2(3a+4)(3a-4)}$ = $\frac{-6a+24}{2(3a+4)(3a-4)}$ = $\frac{12-3a}{(3a+4)(3a-4)}$ = $\frac{3(4-a)}{(3a+4)(3a-4)}$

8) $\frac{6c^{3}}{9-a^{2}}$ · $\frac{a^{2}-6a+9}{4a^{2}c}$ = $\frac{3c^{2}}{(3-a)(3+a)}$ · $\frac{(a-3)^{2}}{2a^{2}}$ = $\frac{3c^{2}}{(3-a)(3+a)}$ · $\frac{(3-a)^{2}}{2a^{2}}$ = $\frac{3c^{2}(3-a)}{(3+a)2a^{2}}$

10) $\frac{10-15b}{(a-b)^2}}$ ÷ $\frac{9b^{2}-4}{3b-3a}$ = $\frac{5(2-3b)}{(b-a)^{2}}$ · $\frac{3(b-a)}{(3b-2)(3b+2)}$ = - $\frac{15(3b-2)}{(b-a)(3b-2)(3b+2)}$ = - $\frac{15}{(b-a)(3b+2)}$ = $\frac{15}{(a-b)(3b+2)}$

( $\frac{x}{x-2}$ - $\frac{x}{x+2}$ - $\frac{x^{2}+4}{4-x^{2}}$ )÷ $\frac{2x+x^{2}}{(2-x)^{2}}$ = ( $\frac{x}{x-2}$ - $\frac{x}{x+2}$ + $\frac{x^{2}+4}{x^{2}-4}$ )· $\frac{(2-x)^{2}}{2x+x^{2}}$ = ( $\frac{x}{x-2}$ - $\frac{x}{x+2}$ + $\frac{x^{2}+4}{(x-2)(x+2)}$ )· $\frac{(x-2)^{2}}{2x+x^{2}}$ = $\frac{x(x+2)-x(x-2)+x^{2}+4}{(x-2)(x+2)}$ · $\frac{(x-2)^{2}}{2x+x^{2}}$ = $\frac{x^{2}+2x-x^{2}+2x+x^{2}+4}{x+2}$ · $\frac{x-2}{2x+x^{2}}$ = $\frac{x^{2}+4x+4}{x+2}$ · $\frac{x-2}{x(2+x)}$ = $\frac{(x+2)^{2}(x-2)}{x(x+2)^{2}}$ = $\frac{x-2}{x}$