ĐKXĐ: \(x\ge\frac{1}{2}\)
- Với \(\frac{1}{2}\le x< 1\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP>0\end{matrix}\right.\) BPT vô nghiệm
- Với \(x\ge1\)
\(\Leftrightarrow\sqrt[3]{7x-8}+1\ge2x-2\sqrt{2x-1}\)
\(\Leftrightarrow x+1-2\sqrt{2x-1}+x-2-\sqrt[3]{7x-8}\le0\)
\(\Leftrightarrow\frac{x^2-6x+5}{\left(x+1\right)^2+2\sqrt{2x-1}}+\frac{x^3-6x^2+5x}{\left(x-2\right)^2+\left(x-2\right)\sqrt[3]{7x-8}+\sqrt[3]{\left(7x-8\right)^2}}\le0\)
\(\Leftrightarrow\left(x^2-6x+5\right)\left(\frac{1}{\left(x+1\right)^2+2\sqrt{2x-1}}+\frac{x}{\left(x-2\right)^2+\left(x-2\right)\sqrt[3]{7x-8}+\sqrt[3]{\left(7x-8\right)^2}}\right)\le0\)
\(\Leftrightarrow x^2-6x+5\le0\) (phần trong ngoặc luôn dương)
\(\Leftrightarrow1\le x\le5\)
