Điều kiện: \(x\geq \frac{3}{2}\)
\((1)\Leftrightarrow 3(x-2)^{2}-(x-1)+\sqrt{2x-3}+(x-1)-\sqrt[3]{3x-5}\leq 0\)
\(\Leftrightarrow 3(x-2)^{2}-\left [ (x-1)-\sqrt{2x-3}\right]+\frac{(x-1)^{3}-(3x-5)}{(x-1)^{2}+(x-1).\sqrt[3]{(3x-5)}+(\sqrt[3]{3x-5})^{2}}\leq 0\)
\(\Leftrightarrow 3(x-2)^{2}-\frac{(x-1)-(2x-3)}{x-1+\sqrt{2x-3}}+\frac{x^{3}-3x^{2}+4}{(x-1)^{2}+(x-1).\sqrt[3]{3x-5}+(\sqrt[3]{3x-5})^{2}}\leq 0\)
Do: \(x\geq \frac{3}{2}\Rightarrow x-1+\sqrt{2x-3}>0\)
Và: \((x-1)^{2}+(x-1)\sqrt[3]{3x-5}+(\sqrt[3]{3x-5})^{2}>0\)
\((1)\Leftrightarrow 3(x-2)^{2}-\frac{(x-2)^{2}}{x-1+\sqrt{2x-3}}+\frac{(x-2)^{2}(x+1)}{(x-1)^{2}+(x-1)\sqrt[3]{3x-5}+(\sqrt[3]{3x-5})}\leq 0\)
\(\Leftrightarrow (x-2)^{2}\left [ 3-\frac{1}{x-1+\sqrt{2x-3}}+\frac{x+1}{(x-1)^{2}+(x-1)\sqrt[3]{3x-5}+(\sqrt[3]{3x-5})} \right ]\leq 0\)
Ta có: \(x\geq \frac{3}{2}\Rightarrow \left\{\begin{matrix} x-1\geq \frac{1}{2}\\\sqrt{2x-3}\geq 0 \end{matrix}\right.\Rightarrow x-1+\sqrt{2x-3}\geq \frac{1}{2}\Rightarrow \frac{1}{x-1+\sqrt{2x-3}}\leq 2\)
\(\Rightarrow 3-\frac{1}{x-1+\sqrt{2x-3}}>0\)
Và \(\frac{x+1}{(x-1)^{2}+(x-1).\sqrt[3]{3x-5}+(\sqrt[3]{(3x-5)})^{2}}>0\; \; \forall x\geq \frac{3}{2}\)
Suy ra bất phương trình \((1)\Leftrightarrow (x-2)\leq 0\Leftrightarrow x=2\) (tmđk)
Vậy tập nghiệm của bất phương trình là S = {2}
