Riješi a/1-a^2+a/1+a^2=

$$\frac { a } { 1 - a ^ { 2 } } + \frac { a } { 1 + a ^ { 2 } } =$$

Procijeni

$$\frac{2a}{1-a^{4}}$$

Kratki koraci rješenja

Faktorirajte 1-a^{2}.

$$\frac{a}{\left(a-1\right)\left(-a-1\right)}+\frac{a}{1+a^{2}}$$

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

$$\frac{a\left(a^{2}+1\right)}{\left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right)}+\frac{a\left(a-1\right)\left(-a-1\right)}{\left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right)}$$

Pošto \frac{a\left(a^{2}+1\right)}{\left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right)} i \frac{a\left(a-1\right)\left(-a-1\right)}{\left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right)} imaju isti imenilac, saberite ih tako što ćete sabrati njihove brojioce.

$$\frac{a\left(a^{2}+1\right)+a\left(a-1\right)\left(-a-1\right)}{\left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right)}$$

Izvršite množenja u a\left(a^{2}+1\right)+a\left(a-1\right)\left(-a-1\right).

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

Kombinirajte slične izraze u a^{3}+a-a^{3}-a^{2}+a^{2}+a.

$$\frac{2a}{\left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right)}$$

Proširite \left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right).

$$\frac{2a}{-a^{4}+1}$$

Faktor

$$\frac{2a}{1-a^{4}}$$