3С.57 3С.58 Пожалуйста!

Решите задачу:

$\sqrt{m^2-2m+1} = \sqrt{(m-1)^2} =|m-1|=1-m, \ m \leq 1 \\\ \sqrt{9m^2-6m+1} = \sqrt{(3m-1)^2} =|3m-1|=1-3m, \ m \ \textless \ \frac{1}{3}$

$\frac{64x^3-27y^6}{9y^4-16x^2} = \frac{(4x-3y^2)(16x^2+12xy^2+9y^4)}{(3y^2-4x)(3y^2+4x)} = - \frac{16x^2+12xy^2+9y^4}{3y^2+4x} \\\ \frac{a^{33}+1}{a^{11}-a^{22}+a^{33}} = \frac{(a^{11}+1)(a^{22}-a^{11}+1)}{a^{11}(1-a^{11}+a^{22})} =\frac{a^{11}+1}{a^{11}}$