Question

Помогите пожалуйста!
Упростите:
4/c^2-9 - 2/c^2+3c

P.S. Только пишите не сразу ответ, о объясните как решали

Answer 1

$\displaystyle\bf\\\frac{4}{c^{2} -9} -\frac{2}{c^{2} +3c} =\frac{4}{(c+3)(c-3)} -\frac{2}{c(c+3)} =\frac{4\cdot c-2\cdot(c-3)}{c(c+3)(c-3)} =\\\\\\=\frac{4c-2c+6}{c(c+3)(c-3)}=\frac{6+2c}{c(c+3)(c-3)}=\frac{2(3+c)}{c(c+3)(c-3)}=\frac{2}{c^{2}-3c }$

Answer 2

Объяснение:

$\frac{4}{c {}^{2} - 9 } - \frac{2}{c {}^{2} + 3c} =\frac{4}{\left(c-3\right)\left(c+3\right)}-\frac{2}{c\left(c+3\right)} =\frac{4c}{c\left(c-3\right)\left(c+3\right)}-\frac{2\left(c-3\right)}{c\left(c-3\right)\left(c+3\right)} =\frac{4c-2\left(c-3\right)}{c\left(c-3\right)\left(c+3\right)} =\frac{4c-2c+6}{c\left(c-3\right)\left(c+3\right)} =\frac{2c+6}{c\left(c-3\right)\left(c+3\right)} =\frac{2\left(c+3\right)}{c\left(c-3\right)\left(c+3\right)} =\frac{2}{c\left(c-3\right)} =\frac{2}{c^{2}-3c}$