Question

помогите пожалуйста!!!
1)
$(\frac{a + 5}{5a-1} + \frac{a + 5}{a + 1} ) \div \frac{ {a}^{2} + 5a } {1 - 5a} + \frac{ {a}^{2} + 5 }{a + 1}$
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Answer 1

Ответ с объяснением:

$(\frac{a+5}{5a-1} + \frac{a+5}{a+1} ) : \frac{a^2+5a}{1-5a} + \frac{a^2+5}{a+1} =$

$=\frac{(a+1)*(a+5)+(5a-1)*(a+5)}{(5a-1)*(a+1)} * \frac{1-5a}{a^2+5a} + \frac{a^2+5}{a+1} = \frac{a^2+5a+a+5+5a^2+25a-a-5}{(5a-1)*(a+1)} * \frac{-(5a-1)}{a^2+5a} + \frac{a^2+5}{a+1} = \frac{a^2+5a+5a^2+25a}{(5a-1)*(a+1)} * \frac{-(5a-1)}{a^2+5a} + \frac{a^2+5}{a+1} = \frac{6a^2+30a}{a+1} * \frac{-1}{a^2+5a} + \frac{a^2+5}{a+1} = \frac{6(a^2+5a)}{a+1} * \frac{-1}{a^2+5a} + \frac{a^2+5}{a+1} = \frac{6}{a+1} * (-1) + \frac{a^2+5}{a+1} = -\frac{6}{a+1} + \frac{a^2+5}{a+1} = \frac{-6+a^2+5}{a+1} =$

$= \frac{-1+a^2}{a+1} = \frac{a^2-1}{a+1} = \frac{(a-1)(a+1)}{a+1} = a-1$

Надеюсь, что помог! <3