agantisbz

2022-07-18

Help to simplify this complicated fraction
I require help to simplify this. I used the method to make the denominator a single fraction then multiply the top by the reciprocal but when it comes to cancelling, I'm not sure if i've done it right.
$\frac{\frac{1}{{x}^{2}+x+2}}{1+\frac{2}{x\left({x}^{2}+x+2\right)}}$

### Answer & Explanation

Killaninl2

Assuming $x\ne 0$ and ${x}^{2}+x+2\ne 0$ (the latter is always the case):
$\frac{\frac{1}{{x}^{2}+x+2}}{1+\frac{2}{x\left({x}^{2}+x+2\right)}}\cdot \frac{x\left({x}^{2}+x+2\right)}{x\left({x}^{2}+x+2\right)}=\frac{x}{x\left({x}^{2}+x+2\right)+2}$
It is not possible to simplify this further, if not by factorizing the denominator. Please keep in mind the conditions I've put in the beginning.
EDIT: factorization leads to the final answer:
$\frac{x}{\left(x+1\right)\left({x}^{2}+2\right)}$
Only for non-zero $x$; now that you've got an explicit form of the fraction, you can put existence conditions. In this case, $x\ne 0$ and $x\ne -1$

jlo2ni5x

$\frac{\frac{1}{{x}^{2}+x+2}}{1+\frac{2}{x\left({x}^{2}+x+2\right)}}=\frac{x}{x\left({x}^{2}+x+2\right)+2}=\frac{x}{\left(x+1\right)•\left({x}^{2}+2\right)}$
where $x\ne -1$

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Recalculate according to your conditions!