Question

Срочно пожалуйста))))

Answer 1

$\frac{16a^2-25c^2}{ac+c^2} :\frac{5c-4a}{3c} +\frac{13a+14c}{a+c} =$

$=\frac{(4a)^2-(5c)^2}{c(a+c)} *\frac{3c}{5c-4a} +\frac{13a+14c}{a+c} =$

$=\frac{(4a-5c)(4a+5c)}{(a+c)} *\frac{3}{5c-4a} +\frac{13a+14c}{a+c} =$

$=-\frac{(5c-4a)(4a+5c)}{(a+c)} *\frac{3}{5c-4a} +\frac{13a+14c}{a+c} =$

$=-\frac{3(4a+5c)}{a+c} +\frac{13a+14c}{a+c} =$

$=\frac{-12a-15c}{a+c} +\frac{13a+14c}{a+c} =$

$=\frac{-12a-15c+13a+14c}{a+c} =\frac{a-c}{a+c}$

Ответ:  $\frac{a-c}{a+c}$