Помогите с алгеброй ПОЖАЛУЙСТА

1) $\frac{a(a+4)+2}{(a+4)(a-4)} = \frac{a^2+4a+2}{(a+4)(a-4)}$
2) $\frac{y(y-3)+6}{(y-3)(y+3)} = \frac{y^2-3y+6}{(y-3)(y+3)}$
3) $\frac{1+b(2b-3)}{(2b-3)(2b+3)} = \frac{2b^2-3b+1}{(2b-3)(2b+3)}$
4) $\frac{1+a(3a+1)}{(3a-1)(3a+1)} = \frac{3a^2+a+1}{(3a-1)(3a+1)}$
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1) $\frac{(y+3)+(y-3)}{(y-3)(y+3)^2} = \frac{2y}{(y-3)(y+3)^2}$
2) $\frac{(a+2)+(a-2)}{(a+2)(a-2)^2} = \frac{2a}{(a+2)(a-2)^2}$
3) $\frac{(2b+5)+(2b-5)}{(2b+5)(2b-5)^2} = \frac{4b}{(2b+5)(2b-5)^2}$
4) $\frac{(3x+4)+(3x-4)}{(3x-4)(3x+4)^2} = \frac{6x}{(3x-4)(3x+4)^2}$
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1) $\frac{1}{2b(3a-b)} + \frac{1}{(b-3a)(b+3a)} + \frac{1}{(3a+b)^2}$ = $\frac{1}{2b(3a-b)} - \frac{1}{(3a-b)(3a+b))} + \frac{1}{(3a+b)^2} =$ =
= $\frac{2b(3a+b)^2-2b(3a+b)+2b(3a-b)}{2b(3a-b)(3a+b)^2}$ =
$\frac{2b(9a+6ab+b^2)-6ab-2b^2+6ab-2b^2}{2b(3a-b)(3a+b)^2}$ =
= $\frac{18ab+12ab^2+2b^3-4b^2}{2b(3a-b)(3a+b)^2}$ = $\frac{2b(9a+6ab+b^2-2b)}{2b(3a-b)(3a+b)^2}$ = $\frac{9a+6ab+b^2-2b}{(3a-b)(3a+b)^2}$