ĐKXĐ: x > 1/4
\(\sqrt{x^3+1}\left(4x-1\right)-2x^3-x^2-1=0\Leftrightarrow\left(\sqrt{x^3+1}-\left(x+1\right)\right)\left(4x-1\right)+\left(x+1\right)\left(4x-1\right)-2x^3-x^2-1=0\Leftrightarrow\left(\sqrt{x^3+1}-\left(x+1\right)\right)\left(4x-1\right)-2x^3+3x^2+3x-2=0\) \(\Leftrightarrow\dfrac{x^3+1-\left(x+1\right)^2}{\sqrt{x^3+1}+x+1}\left(4x-1\right)+\left(2x+1\right)\left(x^2-x-2\right)=0\Leftrightarrow\dfrac{x\left(x^2-x-2\right)}{\sqrt{x^3+1}+x+1}\left(4x-1\right)+\left(2x+1\right)\left(x^2-x-2\right)=0\) \(\Leftrightarrow\left(x^2-x-2\right)\left(\dfrac{x}{\sqrt{x^3+1}+x+1}\left(4x-1\right)+2x+1\right)=0\) (vì theo đkxđ thì \(\dfrac{x}{\sqrt{x^3+1}+x+1}\left(4x-1\right)+2x+1>0\)
Do đó : \(x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Dễ thấy x = -1 thì hai vế của pt bằng nhau và bằng 0 nên ta nhận nghiệm x = -1
Vậy nghiệm của pt là x = -1 ; x = 2
