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.

$\frac{\frac{1}{{x}^{2}+x+2}}{1+\frac{2}{x({x}^{2}+x+2)}}$

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.

$\frac{\frac{1}{{x}^{2}+x+2}}{1+\frac{2}{x({x}^{2}+x+2)}}$

Killaninl2

Beginner2022-07-19Added 20 answers

Assuming $x\ne 0$ and ${x}^{2}+x+2\ne 0$ (the latter is always the case):

$\frac{\frac{1}{{x}^{2}+x+2}}{1+\frac{2}{x({x}^{2}+x+2)}}\cdot \frac{x({x}^{2}+x+2)}{x({x}^{2}+x+2)}=\frac{x}{x({x}^{2}+x+2)+2}$

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:

$\frac{x}{(x+1)({x}^{2}+2)}$

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 $x\ne -1$

$\frac{\frac{1}{{x}^{2}+x+2}}{1+\frac{2}{x({x}^{2}+x+2)}}\cdot \frac{x({x}^{2}+x+2)}{x({x}^{2}+x+2)}=\frac{x}{x({x}^{2}+x+2)+2}$

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:

$\frac{x}{(x+1)({x}^{2}+2)}$

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

Beginner2022-07-20Added 8 answers

$\frac{\frac{1}{{x}^{2}+x+2}}{1+\frac{2}{x({x}^{2}+x+2)}}=\frac{x}{x({x}^{2}+x+2)+2}=\frac{x}{(x+1)\u2022({x}^{2}+2)}$

where $x\ne -1$

where $x\ne -1$

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