Resolver x+1/x^2+x-12-x-4/x^2+7x+12

$$\frac { x + 1 } { x ^ { 2 } + x - 12 } - \frac { x - 4 } { x ^ { 2 } + 7 x + 12 }$$

Calcular

$$\frac{11x-9}{\left(x+4\right)\left(x^{2}-9\right)}$$

Pasos de solución curta

Factoriza x^{2}+x-12. Factoriza x^{2}+7x+12.

$$\frac{x+1}{\left(x-3\right)\left(x+4\right)}-\frac{x-4}{\left(x+3\right)\left(x+4\right)}$$

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

$$\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}-\frac{\left(x-4\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}$$

Dado que \frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)} e \frac{\left(x-4\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)} teñen o mesmo denominador, réstaos mediante a resta dos seus numeradores.

$$\frac{\left(x+1\right)\left(x+3\right)-\left(x-4\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}$$

Fai as multiplicacións en \left(x+1\right)\left(x+3\right)-\left(x-4\right)\left(x-3\right).

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

Combina como termos en x^{2}+3x+x+3-x^{2}+3x+4x-12.

$$\frac{11x-9}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}$$

Expande \left(x-3\right)\left(x+3\right)\left(x+4\right).

$$\frac{11x-9}{x^{3}+4x^{2}-9x-36}$$

Expandir

$$\frac{11x-9}{\left(x+4\right)\left(x^{2}-9\right)}$$

Pasos de solución curta

Factoriza x^{2}+x-12. Factoriza x^{2}+7x+12.

$$\frac{x+1}{\left(x-3\right)\left(x+4\right)}-\frac{x-4}{\left(x+3\right)\left(x+4\right)}$$

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

$$\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}-\frac{\left(x-4\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}$$

Dado que \frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)} e \frac{\left(x-4\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)} teñen o mesmo denominador, réstaos mediante a resta dos seus numeradores.

$$\frac{\left(x+1\right)\left(x+3\right)-\left(x-4\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}$$

Fai as multiplicacións en \left(x+1\right)\left(x+3\right)-\left(x-4\right)\left(x-3\right).

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

Combina como termos en x^{2}+3x+x+3-x^{2}+3x+4x-12.

$$\frac{11x-9}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}$$

Expande \left(x-3\right)\left(x+3\right)\left(x+4\right).

$$\frac{11x-9}{x^{3}+4x^{2}-9x-36}$$