Riješi 1/2x^2-x-3-1/2x^2+x-1

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

Procijeni

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

Kratki koraci rješenja

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

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

Da biste izvršili zbrajanje ili oduzimanje izraza, rastavite ih kako bi im nazivnici bili isti. Najmanji zajednički množilac brojeva \left(2x-3\right)\left(x+1\right) i \left(2x-1\right)\left(x+1\right) je \left(2x-3\right)\left(2x-1\right)\left(x+1\right). Pomnožite \frac{1}{\left(2x-3\right)\left(x+1\right)} i \frac{2x-1}{2x-1}. Pomnožite \frac{1}{\left(2x-1\right)\left(x+1\right)} i \frac{2x-3}{2x-3}.

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

Pošto \frac{2x-1}{\left(2x-3\right)\left(2x-1\right)\left(x+1\right)} i \frac{2x-3}{\left(2x-3\right)\left(2x-1\right)\left(x+1\right)} imaju isti imenilac, oduzmite ih tako što ćete oduzeti njihove brojioce.

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

Izvršite množenja u 2x-1-\left(2x-3\right).

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

Kombinirajte slične izraze u 2x-1-2x+3.

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

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

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

Razlikovanje u pogledu x

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