Question

Выполни сложение алгебраических дробей d+3/(3−d)2+3/3d−d2.

Answer 1

Ответ:

$\frac{d {}^{2} + 9 }{d( 3 - d) {}^{2} }$

Объяснение:

$\frac{d + 3}{(3 - d) {}^{2} } + \frac{3}{3d - d {}^{2} } =\frac{d+3}{\left(3-d\right)^{2}}+\frac{3}{d\left(-d+3\right)} =\frac{\left(d+3\right)d\left(-d+3\right)}{d\left(-d+3\right)\left(-d+3\right)^{2}}+\frac{3\left(-d+3\right)^{2}}{d\left(-d+3\right)\left(-d+3\right)^{2}} =\frac{\left(d+3\right)d\left(-d+3\right)+3\left(-d+3\right)^{2}}{d\left(-d+3\right)\left(-d+3\right)^{2}} =\frac{-d^{3}+3d^{2}-3d^{2}+9d+3d^{2}-18d+27}{d\left(-d+3\right)\left(-d+3\right)^{2}} =\frac{-9d+27+3d^{2}-d^{3}}{d\left(-d+3\right)\left(-d+3\right)^{2}} =\frac{\left(-d+3\right)\left(d^{2}+9\right)}{d\left(-d+3\right)\left(-d+3\right)^{2}} =\frac{d^{2}+9}{d\left(-d+3\right)^{2}} =\frac{d^{2}+9}{d^{3}-6d^{2}+9d} =\frac{d^{2}+9}{d\left(3-d\right)^{2}}$