Question

Нужно решить номер 13, 14. Заранее СПС.

Answer 1

$\frac{5y - 4}{6y} + \frac{y + 2}{3y} = \frac{5y - 4}{6y} + \frac{2(y + 2)}{2 \cdot 3y} = \\ = \frac{5y - 4 + 2y + 4}{6y} = \frac{7y}{6y} = \frac{7}{6}$
$\frac{b + 2}{15b} - \frac{3c - 5}{30c} = \\ = \frac{2c(b + 2)}{2c \cdot 15b} - \frac{(3c - 5)b}{30c \cdot b} = \\ = \frac{2bc + 4c - 3bc + 5b}{30bc} = \\ = \frac{5b + 4c - bc}{30bc}$
$\frac{2a + b}{2 {a}^{2} - ab} - \frac{16a}{4 {a}^{2} - {b}^{2} } - \frac{2a - b}{2 {a}^{2} + ab} = \\ = \frac{2a + b}{a( 2{a} - b)} - \frac{16a}{(2a - b)(2a + b) } - \\ - \frac{2a - b}{a(2 {a} + b)} = \\ = \frac{(2a + b) ^{2} - 16 {a}^{2} - (2a - b) ^{2} }{a(2a - b)(2a + b)} = \\ = \frac{(2a + b - 2a + b)(2a + b + 2a - b) - 16 {a}^{2} }{a(2a - b)(2a + b)} = \\ = \frac{2b \cdot4a - 16 {a}^{2} }{a(2a - b)(2a + b)} = \\ = \frac{ - 8a(2a - b)}{a(2a - b)(2a + b)} = - \frac{8}{2a + b}$