besnuffelfo

2022-09-24

How to simplify this fraction with different powers?
I happen to be stuck trying to simplify this:
$\left[\frac{\left(3x+2\right)\left(x+1{\right)}^{\frac{3}{2}}-\left(\frac{3}{2}{x}^{2}+2x\right)\left(\frac{3}{2}\right)\left(x+1{\right)}^{\frac{1}{2}}}{\left(x+1{\right)}^{3}}\right]$

Do you have a similar question?

kregde84

Expert

We have
$\frac{\left(3x+2\right)\left(x+1{\right)}^{\frac{3}{2}}-\left(\frac{3}{2}{x}^{2}+2x\right)\left(\frac{3}{2}\right)\left(x+1{\right)}^{\frac{1}{2}}}{\left(x+1{\right)}^{3}}$
divide numerator and denominator by $\left(x+1{\right)}^{\frac{1}{2}}$ giving
$\frac{\left(3x+2\right)\left(x+1\right)-\left(\frac{3}{2}{x}^{2}+2x\right)\left(\frac{3}{2}\right)}{\left(x+1{\right)}^{\frac{5}{2}}}$
multiply the $\left(3x+2\right)\left(x+1\right)$ to get $3{x}^{2}+5x+2$ and the $\left(\frac{3}{2}{x}^{2}+2x\right)\left(\frac{3}{2}\right)$ to get $\frac{9}{4}{x}^{2}+3x$ giving:
$\frac{3{x}^{2}+5x+2-\frac{9}{4}{x}^{2}-3x}{\left(x+1{\right)}^{\frac{5}{2}}}$
then simplify the numerator:
$3{x}^{2}+5x+2-\frac{9}{4}{x}^{2}-3x=\frac{12}{4}{x}^{2}-\frac{9}{4}{x}^{2}+5x-3x+2=\frac{3}{4}{x}^{2}+2x+2$
to get
$\frac{\frac{3}{4}{x}^{2}+2x+2}{\left(x+1{\right)}^{\frac{5}{2}}}$

Still Have Questions?

demitereur

Expert

$\frac{\left(3x+2\right)\left(x+1{\right)}^{\frac{3}{2}}-\left(\frac{3}{2}{x}^{2}+2x\right)\left(\frac{3}{2}\right)\left(x+1{\right)}^{\frac{1}{2}}}{\left(x+1{\right)}^{3}}=$
take the common factor $\left(x+1{\right)}^{\frac{1}{2}}$
$=\frac{\left[\left(3x+2\right)\left(x+1\right)-\left(\frac{3}{2}{x}^{2}+2x\right)\left(\frac{3}{2}\right)\right]\left(x+1{\right)}^{\frac{1}{2}}}{\left(x+1{\right)}^{3}}=$
$=\frac{\left(3x+2\right)\left(x+1\right)-\left(\frac{3}{2}{x}^{2}+2x\right)\left(\frac{3}{2}\right)}{\left(x+1{\right)}^{3}\left(x+1{\right)}^{-\frac{1}{2}}}=\frac{\left(3/4\right){x}^{2}+2x+2}{\left(x+1{\right)}^{\frac{5}{2}}}$

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