Помогите,пожалуйста решить 2;3;4.Очень нужно

$........=\frac{a(a-3b)}{5} *\frac{10b}{a^{2}} +\frac{6b^{2}}{a} =\frac{2b(a-3b)}{a}+ \frac{6b^{2}}{a} =\frac{2ab-6b^{2}+6b^{2}}{a} =2b$

$........=\frac{(x-5)(x+5)}{x+6}* \frac{1}{x(x+5)}- \frac{x+5}{x(x-6)} =\frac{x-5}{x(x+6)} -\frac{x+5}{x(x-6)} =\frac{(x-5)(x-6)-(x+5)(x+6)}{x(x+6)(x-6)} =\\ \\ \frac{x^{2}-11x+30-x^{2}-11x-30}{x(x+6)(x-6)} =-\frac{22x}{x(x+6)(x-6)} =\frac{22}{x^{2}-36}$

$...........=\frac{x+3}{x^{2}+9}* \frac{(x+3)(x+3)+(x-3)(x-3)}{(x-3)(x+3)} =\frac{x+3}{x^{2}+9}* \frac{x^{2}+6x+9+x^{2}-6x+9}{(x-3)(x+3)} =\\ \\ \frac{2x^{2}+18}{(x^{2}+9)(x-3)} =\frac{2(x^{2}+9)}{(x^{2}+9)(x-3)} =\frac{2}{x-3} \\ \\ x=1.75\\ \\ \frac{2}{1.75-3} =-\frac{2}{1.25}=-1.6$

$.........=\frac{24x-16x^{2}-9}{24x} *\frac{12x}{3-4x} -2x=\frac{-(9-24x+16x^{2})}{2(3-4x)} -2x=\\ \\ -\frac{(3-4x)(3-4x)}{2(3-4x)}-2x= -\frac{3-4x-4x}{2} =-\frac{3-8x}{2}$