Resolver 1/2x^2-x-3-1/2x^2+x-1

$$\frac { 1 } { 2 x ^ { 2 } - x - 3 } - \frac { 1 } { 2 x ^ { 2 } + x - 1 }$$

Calcular

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

Pasos de solución curta

Factoriza 2x^{2}-x-3. Factoriza 2x^{2}+x-1.

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

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

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

Dado que \frac{2x-1}{\left(2x-3\right)\left(2x-1\right)\left(x+1\right)} e \frac{2x-3}{\left(2x-3\right)\left(2x-1\right)\left(x+1\right)} teñen o mesmo denominador, réstaos mediante a resta dos seus numeradores.

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

Fai as multiplicacións en 2x-1-\left(2x-3\right).

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

Combina como termos en 2x-1-2x+3.

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

Expande \left(2x-3\right)\left(2x-1\right)\left(x+1\right).

$$\frac{2}{4x^{3}-4x^{2}-5x+3}$$

Diferenciar w.r.t. x

$$-\frac{2\left(\left(6x-2\right)^{2}-19\right)}{3\left(\left(2x-3\right)\left(2x-1\right)\left(x+1\right)\right)^{2}}$$