Question

ПОМОЩЬ ПОМОЩЬ ДАМ МНОГО БАЛЛОВ

Answer 1

Ответ:

а)

$\frac{3b {}^{2} + 2b }{b {}^{2} - 4} - \frac{b}{b - 2} =\frac{3b^{2}+2b}{\left(b-2\right)\left(b+2\right)}-\frac{b}{b-2} =\frac{3b^{2}+2b}{\left(b-2\right)\left(b+2\right)}-\frac{b\left(b+2\right)}{\left(b-2\right)\left(b+2\right)} =\frac{3b^{2}+2b-b\left(b+2\right)}{\left(b-2\right)\left(b+2\right)} =\frac{3b^{2}+2b-b^{2}-2b}{\left(b-2\right)\left(b+2\right)} =\frac{2b^{2}}{\left(b-2\right)\left(b+2\right)} =\frac{2b^{2}}{b^{2}-4}$

б)

$\frac{2 + 5c {}^{2} }{c} - 6c = \frac{2+5c^{2}}{c}+\frac{-6cc}{c} =\frac{2+5c^{2}-6cc}{c} =\frac{2+5c^{2}-6c^{2}}{c} =\frac{2-c^{2}}{c}$

в)

$\frac{xy + y {}^{2} }{8x} \div \frac{x + y}{2x} = \frac{(xy + y {}^{2} ) \times 2x}{8x(x + y)} = \frac{xy + y {}^{2} }{4(x + y)} = \frac{y(x + y)}{4(x + y)} = \frac{y}{4}$