Алгебра . Помогите, с решением.

Решите задачу:

$(a^{-5})^4*a^{22} = a^{-20}*a^{22} =a^2\\0,4x^6y^{-8}*50x^{-5}y^9= 20x^{6-5}y^{-8+9}=xy$

$(\frac{1}{6}x^{-4}y^3)^{-1} = \frac{6x^4}{y^3}\\(\frac{3a^{-4}}{2b^{-3}})^{-2} * 10a^7b^3 = \frac{4b^{-6}}{9a^{-8}}* 10a^7b^3 = \frac{4b^{-6}}{9}* 10a^{15}b^3 = \frac{40a^{15}}{9b^3}$

$\frac{2^{-6}*4^{-3}}{8^{-7}} = \frac{2^{-6}*2^{-6}}{2^{-21}} = 2^{15}*2^{-6} = 2^9 = 512$

$(3,5*10^{-5})(6,4*10^2) = (3,5*6,4)(10^{-5}*10^2) = 22,4*10^{-3}$

$(x^{-1}-y^{-1})(x-y)^{-1} = (\frac{1}{x}-y)*\frac{1}{x-y} = \frac{1-xy}{x}*\frac{1}{x-y}=\frac{1-xy}{x(x-y)} = \frac{1-xy}{x^2-xy}$