ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\x\ge\dfrac{3}{5}\end{matrix}\right.\)
\(\left(x+1\right)\left(45x^2-62x+25\right)=4\sqrt{\left(x+1\right)\left(5x-3\right)\left(5x-3\right)^2}\)
- Với \(x=-1\) là 1 nghiệm
- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP>0\end{matrix}\right.\) pt vô nghiệm
Với \(x\ge\dfrac{3}{5}\) ta có:
\(45x^3-17x^2-37x+25=4\sqrt{\left(x+1\right)\left(5x-3\right)\left(5x-3\right)^2}\)
\(\Leftrightarrow45x^3-17x^2-37x+25\le2\left[\left(x+1\right)\left(5x-3\right)+\left(5x-3\right)^2\right]\)
\(\Leftrightarrow45x^3-77x^2+19x+13\le0\)
\(\Leftrightarrow\left(x-1\right)^2\left(45x+13\right)\le0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
