Помогите , решить алгебру

Question 1

Question 2

1)
$\frac{x^4 + 1}{x^7 +x^3} = \frac{x^4+1}{x^3*(x^4 + 1)} = \frac{1}{x^3} \\ \\ x= - \frac{1}{3} \\ \frac{1}{(- \frac{1}{3})^3 } = 1 : (- \frac{1}{27} ) = 1 * (- \frac{27}{1} ) = -27$
ответ: - 27

2)
$\frac{a+4}{a^7} - \frac{4a^4 +1}{a^{11} } = \frac{a+4}{a^7} - \frac{4a^4+1}{a^4*a^7} = \frac{a^4(a+4) -(4a^4+1)}{a^{11}} = \\ \\ = \frac{a^5 +4a^4 - 4a^4 - 1}{a^{11}} = \frac{a^5 -1}{a^{11}}$
ответ: 4) $\frac{a^5 -1}{a^{11}}$

3)
$\frac{2x^2}{3x-6y} : \frac{x}{x-2y} = \frac{2x^2}{3*(x-2y)} * \frac{x-2y}{x} = \frac{2x*x*(x-2y)}{3*x*(x-2y)} = \frac{2x}{3}$
ответ: 1) $\frac{2x}{3}$

4)
$( \frac{2p^2}{k^3} )^3 : \frac{4p^5}{k^{10}} = \frac{(2^1p^2)^3}{(k^3)^3} * \frac{k^{10}}{2^2*p^5} = \frac{2^{1*3} *p^{2*3}*k^{10}}{k^{3*3} *2^2*p^5} = \\ \\ = \frac{2^3*p^6*k^{10}}{2^2*p^5*k^9} = 2^{3-2} *p^{6-5} * k^{10-9} = 2^1*p^1*k^1= 2pk \\ \\ p=1.5 ; k=3.2 \\ 2 * 1.5 * 3.2 = 3 * 3.2 = 9.6$
ответ: 9,6 .

5)
$\frac{c}{c-5} + \frac{c^2}{(c-5)^2} * \frac{25-c^2}{5c+25} = \frac{c}{c-5} + \frac{c^2 * ( - (c^2 - 5^2) )}{(c-5)^2 * 5*(c+5)} = \\ \\ =\frac{c}{c-5} + \frac{- c^2 * (c-5)(c+5)}{5(c-5)(c-5)(c+5)} = \frac{c}{c-5} + \frac{-c^2}{5(c-5)} = \\ \\ =\frac{c}{c-5} - \frac{c^2}{5(c-5)} = \frac{5c-c^2}{5(c-5)} = \frac{-c(c-5)}{5(c-5)} = \\ \\ = \frac{-c}{5}= - \frac{c}{5}$
ответ : 3) $- \frac{c}{5}$