ĐK: $x\geq \frac{1}{2}$BPT$\Leftrightarrow 2x-1-2\sqrt{2x-1}-\sqrt[3]{7x-8}\leq 0$
$\Leftrightarrow [(x+1)-2\sqrt{2x-1}]+[(x-2)-\sqrt[3]{7x-8}]\leq 0$
$\Leftrightarrow \frac{(x+1)^{2}-4(2x-1)}{x+1+2\sqrt{2x-1}}+\frac{(x-2)^{3}-(7x-8)}{(x-2)^{2}+(x-2)\sqrt[3]{7x-8}+\sqrt[3]{(7x-8)^{2}}}\leq 0$
$\Leftrightarrow \frac{(x-1)(x-5)}{...........}+\frac{x(x-1)(x-5)}{............}\leq 0$
$\Leftrightarrow (x-1)(x-5)(\frac{1}{x+1+2\sqrt{2x-1}}+\frac{x}{(x-2)^{2}+(x-2)\sqrt[3]{7x-8}+\sqrt[3]{(7x-8)^{2}}})\leq 0$
Mà $(......)>0\Rightarrow (x-1)(x-5)\leq 0\Rightarrow 1\leq x\leq 5$
Vậy $1\leq x\leq 5$
Xem có đúng không nếu đúng thì vote + Tích V cho mình nha