Предмет: Алгебра, автор: chel1189

ПОМОГИТЕ ПОЖАЛУЙСТА!!!!

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Ответы

Автор ответа: svit111

Ответ:

$\frac{ 2 }{ 3+b } - \frac{ 12 }{ b ^ { 2 } -9 } + \frac{ 12 }{ 3-b } = \frac{2}{3+b}-\frac{12}{\left(b-3\right)\left(b+3\right)}+\frac{12}{3-b} = \frac{2\left(b-3\right)}{\left(b-3\right)\left(b+3\right)}-\frac{12}{\left(b-3\right)\left(b+3\right)}+\frac{12}{3-b} = \frac{2\left(b-3\right)-12}{\left(b-3\right)\left(b+3\right)}+\frac{12}{3-b} = \frac{2b-6-12}{\left(b-3\right)\left(b+3\right)}+\frac{12}{3-b} = \frac{2b-18}{\left(b-3\right)\left(b+3\right)}+\frac{12}{3-b} = \frac{2b-18}{\left(b-3\right)\left(b+3\right)}+\frac{12\left(-1\right)\left(b+3\right)}{\left(b-3\right)\left(b+3\right)} = \frac{2b-18+12\left(-1\right)\left(b+3\right)}{\left(b-3\right)\left(b+3\right)} =\frac{2b-18-12b-36}{\left(b-3\right)\left(b+3\right)} = \frac{-10b-54}{\left(b-3\right)\left(b+3\right)} = \frac{-10b-54}{b^{2}-9}$