Resolver 3/3+b-12-b/b^2-b-6-2/2-6

$$\frac{ 3 }{ 3+b } - \frac{ 12-b }{ { b }^{ 2 } -b-6 } - \frac{ 2 }{ 2-6 }$$

Calcular

$$\frac{b^{3}+10b^{2}-33b-126}{2\left(b+2\right)\left(b^{2}-9\right)}$$

Pasos de solución curta

Resta 6 de 2 para obter -4.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}-\frac{2}{-4}$$

Reduce a fracción \frac{2}{-4} a termos máis baixos extraendo e cancelando 2.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}-\left(-\frac{1}{2}\right)$$

O contrario de -\frac{1}{2} é \frac{1}{2}.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}+\frac{1}{2}$$

Factoriza b^{2}-b-6.

$$\frac{3}{3+b}-\frac{12-b}{\left(b-3\right)\left(b+2\right)}+\frac{1}{2}$$

Para sumar ou restar expresións, expándeas para facer que os seus denominadores sexan iguais. O mínimo común múltiplo de 3+b e \left(b-3\right)\left(b+2\right) é \left(b-3\right)\left(b+2\right)\left(b+3\right). Multiplica \frac{3}{3+b} por \frac{\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)}. Multiplica \frac{12-b}{\left(b-3\right)\left(b+2\right)} por \frac{b+3}{b+3}.

$$\frac{3\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}-\frac{\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Dado que \frac{3\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} e \frac{\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} teñen o mesmo denominador, réstaos mediante a resta dos seus numeradores.

$$\frac{3\left(b-3\right)\left(b+2\right)-\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Fai as multiplicacións en 3\left(b-3\right)\left(b+2\right)-\left(12-b\right)\left(b+3\right).

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

Combina como termos en 3b^{2}+6b-9b-18-12b-36+b^{2}+3b.

$$\frac{4b^{2}-12b-54}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Para sumar ou restar expresións, expándeas para facer que os seus denominadores sexan iguais. O mínimo común múltiplo de \left(b-3\right)\left(b+2\right)\left(b+3\right) e 2 é 2\left(b-3\right)\left(b+2\right)\left(b+3\right). Multiplica \frac{4b^{2}-12b-54}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} por \frac{2}{2}. Multiplica \frac{1}{2} por \frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}.

$$\frac{2\left(4b^{2}-12b-54\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Dado que \frac{2\left(4b^{2}-12b-54\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)} e \frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)} teñen o mesmo denominador, súmaos mediante a suma dos seus numeradores.

$$\frac{2\left(4b^{2}-12b-54\right)+\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Fai as multiplicacións en 2\left(4b^{2}-12b-54\right)+\left(b-3\right)\left(b+2\right)\left(b+3\right).

$$\frac{8b^{2}-24b-108+b^{3}+5b^{2}+6b-3b^{2}-15b-18}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Combina como termos en 8b^{2}-24b-108+b^{3}+5b^{2}+6b-3b^{2}-15b-18.

$$\frac{10b^{2}-33b-126+b^{3}}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Expande 2\left(b-3\right)\left(b+2\right)\left(b+3\right).

$$\frac{10b^{2}-33b-126+b^{3}}{2b^{3}+4b^{2}-18b-36}$$

Expandir

$$\frac{b^{3}+10b^{2}-33b-126}{2\left(b+3\right)\left(b^{2}-b-6\right)}$$

Pasos de solución curta

Resta 6 de 2 para obter -4.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}-\frac{2}{-4}$$

Reduce a fracción \frac{2}{-4} a termos máis baixos extraendo e cancelando 2.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}-\left(-\frac{1}{2}\right)$$

O contrario de -\frac{1}{2} é \frac{1}{2}.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}+\frac{1}{2}$$

Factoriza b^{2}-b-6.

$$\frac{3}{3+b}-\frac{12-b}{\left(b-3\right)\left(b+2\right)}+\frac{1}{2}$$

Para sumar ou restar expresións, expándeas para facer que os seus denominadores sexan iguais. O mínimo común múltiplo de 3+b e \left(b-3\right)\left(b+2\right) é \left(b-3\right)\left(b+2\right)\left(b+3\right). Multiplica \frac{3}{3+b} por \frac{\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)}. Multiplica \frac{12-b}{\left(b-3\right)\left(b+2\right)} por \frac{b+3}{b+3}.

$$\frac{3\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}-\frac{\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Dado que \frac{3\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} e \frac{\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} teñen o mesmo denominador, réstaos mediante a resta dos seus numeradores.

$$\frac{3\left(b-3\right)\left(b+2\right)-\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Fai as multiplicacións en 3\left(b-3\right)\left(b+2\right)-\left(12-b\right)\left(b+3\right).

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

Combina como termos en 3b^{2}+6b-9b-18-12b-36+b^{2}+3b.

$$\frac{4b^{2}-12b-54}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Para sumar ou restar expresións, expándeas para facer que os seus denominadores sexan iguais. O mínimo común múltiplo de \left(b-3\right)\left(b+2\right)\left(b+3\right) e 2 é 2\left(b-3\right)\left(b+2\right)\left(b+3\right). Multiplica \frac{4b^{2}-12b-54}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} por \frac{2}{2}. Multiplica \frac{1}{2} por \frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}.

$$\frac{2\left(4b^{2}-12b-54\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Dado que \frac{2\left(4b^{2}-12b-54\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)} e \frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)} teñen o mesmo denominador, súmaos mediante a suma dos seus numeradores.

$$\frac{2\left(4b^{2}-12b-54\right)+\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Fai as multiplicacións en 2\left(4b^{2}-12b-54\right)+\left(b-3\right)\left(b+2\right)\left(b+3\right).

$$\frac{8b^{2}-24b-108+b^{3}+5b^{2}+6b-3b^{2}-15b-18}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Combina como termos en 8b^{2}-24b-108+b^{3}+5b^{2}+6b-3b^{2}-15b-18.

$$\frac{10b^{2}-33b-126+b^{3}}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Expande 2\left(b-3\right)\left(b+2\right)\left(b+3\right).

$$\frac{10b^{2}-33b-126+b^{3}}{2b^{3}+4b^{2}-18b-36}$$