Rešite a/1-a^2+a/1+a^2=

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

Ovrednoti

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

Kratka navodila za rešitev

Faktorizirajte 1-a^{2}.

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

Če želite prišteti ali odšteti izraze, jih razširite na skupne imenovalce. Najmanjši skupni mnogokratnik \left(a-1\right)\left(-a-1\right) in 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)} s/z \frac{a^{2}+1}{a^{2}+1}. Pomnožite \frac{a}{1+a^{2}} s/z \frac{\left(a-1\right)\left(-a-1\right)}{\left(a-1\right)\left(-a-1\right)}.

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

\frac{a\left(a^{2}+1\right)}{\left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right)} in \frac{a\left(a-1\right)\left(-a-1\right)}{\left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right)} imata isti imenovalec, zato ju seštejte tako, da seštejete njuna števca.

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

Izvedi množenje v a\left(a^{2}+1\right)+a\left(a-1\right)\left(-a-1\right).

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

Združite podobne člene v a^{3}+a-a^{3}-a^{2}+a^{2}+a.

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

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

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

Faktoriziraj

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