Question

Разложить на множители

8+3с
_____ =
9с²-64

36-y²
_____ =
6-y

100-49d²
_________ =
7d+10

Answer 1

$\frac{8+3c}{(3c+8)(3c-8)} = \frac{1}{3c-8}$
$\frac{(6-y)(6+y)}{6-y} =6+y$
$\frac{(10-7d)(10+7d)}{7d+10} =10-7d$