Question

Помогите пожалуйста, решила, но ответ не совпадает(

Answer 1

Решение внизу на фото

Answer 2

$1) \frac{4c +3y }{ {c}^{2} - {y}^{2} } - \frac{3}{c - y} = \frac{4c + 3y}{(c - y)(c + y)} - \frac{3(c + y)}{(c - y)(c + y)} = \frac{4c + 3y - 3c - 3y}{ (c - y)(c + y)} = \frac{c}{(c - y)(c + y)}$
$2) \: \: \frac{c}{(c - y)(c + y)} + \frac{1}{y - c} = \frac{c}{(c - y)(c + y)} - \frac{1}{c - y} = \frac{c}{(c - y)(c + y)} - \frac{c + y}{(c - y)(c + y)} = \frac{c - c - y}{(c - y)(c + y)} = \frac{ - y}{(c - y)(c + y)}$

$\frac{ - y}{(c - y)(c + y)} - \frac{4}{5c + 5y} = \frac{ - y}{(c - y)(c + y)} - \frac{4}{5(c + y)} = \frac{ - 5y}{5(c - y)(c + y )} - \frac{4(c - y)}{5(c - y)(c + y)} = \frac{ - 5y - 4c + 4y}{5( {c}^{2} - {y}^{2}) } = \frac{ -4c - 5y}{5( {c}^{2} - {y}^{2}) } = \frac{ - (4c + y)}{ - 5( {y}^{2} - {c}^{2}) } = \frac{4c + y}{5( {y}^{2} - {c}^{2} )}$