Помогите решить пожалуйста, алгебра 7 класс , спасибо заранее

Question 1

Question 2

№1.
$2^{-2} * 2^{-2} = 2^{-2 +(-2) } = 2^{-4} = \frac{1}{2^4} = \frac{1}{16} \\ \\ \\ \frac{2^{-3} *2^{-7}}{2^{-5} * 2^{-4}} = \frac{2^{-10}}{2^{-9}} =2^{-10-(-9)}=2^{-1} = \frac{1}{2^1} = 0.5 \\ \\ \\ \frac{(7^{-3})^{-2} * 7^{-5}}{7^{-2} * 7^3} = \frac{7^{-3*(-2)} * 7^{-5}}{7^{-2+3}} = \frac{7^{6+ (-5)}}{7^1} = \frac{7^1}{7^1} = 1 \\ \\ \\ ( \frac{1}{5} )^{-2} * ( \frac{1}{5} )^3 = (5^{-1})^{-2} * (5^{-1})^3 = 5^{-1*(-2) + (-1)*3} = \\ \\ = 5^{-1} = \frac{1}{5} =0.2$

№2.
$а^{-3} = \frac{1}{a^3} \\ \\ (5x)^{-2} = \frac{1}{(5x)^2} = \frac{1}{25x^2} \\ \\ xy^{-1} = x * \frac{1}{y^1} = \frac{x}{1} * \frac{1}{y}= \frac{x}{y} \\ \\ 3m^2n^{-2} = 3m^2 * \frac{1}{n^2} = \frac{3m^2}{n^2} \\ \\ a^{-2} +b^{-2} = \frac{1}{a^2} + \frac{1}{b^2} = \frac{b^2+a^2}{a^2b^2} \\ \\ (u-v)^{-2} = \frac{1}{(u-v)^2} \\ \\ -10yz^{-19} =-10y * \frac{1}{z^{19}} = - \frac{10y}{z^{19}} \\ \\ 2(a+c)^{-3} = \frac{2}{(a+c)^3}$

№3.
$\frac{2}{3^{-17}} = 2 : \frac{1}{3^{17}} = 2 * \frac{3^{17}}{1} = 2*3^{17} \\ \\ \frac{a^{-2}}{b^{-3}} = \frac{1}{a^2} : \frac{1}{b^{3}} = \frac{1}{a^2}* \frac{b^3}{1} = \frac{b^3}{a^2} \\ \\ \frac{m}{np^{-2}} = \frac{mp^2}{n} \\ \\ \frac{ab}{(a+b)^{-2}} = \frac{ab*(a+b)^2}{1} = ab(a+b)^2 \\ \\ \frac{xy}{z^{-1}} = xyz^1 =xyz \\ \\ \frac{z}{x^ny^{-k}} = \frac{zy^k}{x^n}$