floymdiT

2021-11-10

To calculate the expression $3x{\left({x}^{2}+1\right)}^{\frac{1}{2}}\left(2{x}^{2}-x\right)+2{\left({x}^{2}+1\right)}^{\frac{2}{3}}\left(4x-1\right)$

Pohanginah

Calculation:
Given $3x{\left({x}^{2}+1\right)}^{\frac{1}{2}}\left(2{x}^{2}-x\right)+2{\left({x}^{2}+1\right)}^{\frac{2}{3}}\left(4x-1\right)$
This can be written as
$3x{\left({x}^{2}+1\right)}^{\frac{1}{2}}\left(2{x}^{2}-x\right)+2{\left({x}^{2}+1\right)}^{\frac{3}{2}}\left(4x-1\right)=3\cdot x\cdot {\left({x}^{2}+1\right)}^{\frac{1}{2}}\cdot \left(2{x}^{2}-x\right)+2\cdot {\left({x}^{2}+1\right)}^{\frac{1}{2}}\cdot \left({x}^{2}+1\right)\cdot \left(4x-1\right)$
Factoring out the common terms,
$3x{\left({x}^{2}+1\right)}^{\frac{1}{2}}\left(2{x}^{2}-x\right)+2{\left({x}^{2}+1\right)}^{\frac{3}{2}}\left(4x-1\right)$
$={\left({x}^{2}+1\right)}^{\frac{1}{2}}\left(3\cdot x\cdot \left(2{x}^{2}-x\right)+2\cdot \left({x}^{2}+1\right)\cdot \left(4x-1\right)\right)$
$3x{\left({x}^{2}+1\right)}^{\frac{1}{2}}\left(2{x}^{2}-x\right)+2{\left({x}^{2}+1\right)}^{\frac{3}{2}}\left(4x-1\right)$
$={\left({x}^{2}+1\right)}^{\frac{1}{2}}\left(3\cdot x\cdot 2{x}^{2}-3\cdot x\cdot x+2\cdot {x}^{2}\cdot 4x-2\cdot {x}^{2}\cdot 1+2\cdot 1\cdot 4x-2\cdot 1\cdot 1\right)$
$3x{\left({x}^{2}+1\right)}^{\frac{1}{2}}\left(2{x}^{2}-x\right)+2{\left({x}^{2}+1\right)}^{\frac{3}{2}}\left(4x-1\right)$

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