đặt: $\sqrt{3-x}=a; \sqrt{x-1}=b$
bất phương trình tương đương:
$2a-4b+ab-a^2+2b^2\leq 0$
<=> $(b-\frac{a+1}{2})(b-\frac{6-4a}{4})\leq 0$
$<=> (2b-a-1)(4b+4a-6)\leq 0$
* TH1: $\left\{\begin{matrix} 2b-a-1\leq 0\\ a+b\geq \frac{3}{2} \end{matrix}\right.$
<=> $\left\{\begin{matrix} 2\sqrt{x-1}-\sqrt{3-x}-1\leq 0\\ \sqrt{x-1}+\sqrt{3-x}\geq \frac{3}{2} \end{matrix}\right.$
Có: $\sqrt{x-1}+\sqrt{3-x}\geq \frac{3}{2} <=> 2+2\sqrt{(x-1)(3-x)}\geq \frac{3}{2} <=> 2\sqrt{(x-1)(3-x)}\geq \frac{-1}{2}$ (luôn đúng)
$2\sqrt{x-1}-\sqrt{3-x}-1\leq 0 <=> 2\sqrt{x-1}\leq \sqrt{3-x}+1 <=> 4x-4\leq 4-x+2\sqrt{3-x} <=> 5x-8\leq 2\sqrt{3-x}$
==> dễ giải bpt cuối
* TH2: $\left\{\begin{matrix} 2b-a-1\geqslant 0\\a+b\leq \frac{3}{2} \end{matrix}\right.$
<=> $\left\{\begin{matrix} 2\sqrt{x-1}-\sqrt{3-x}-1\geqslant 0\\\sqrt{3-x}+\sqrt{x-1}\leq \frac{3}{2} \end{matrix}\right.$
mà có: $\sqrt{x-1}+\sqrt{3-x}\leq\frac{3}{2} <=> 2+2\sqrt{(x-1)(3-x)}\leg \frac{3}{2} <=> 2\sqrt{(x-1)(3-x)}\leg \frac{-1}{2}$ (vô lý)
Vậy . . . .