Lời giải:
ĐKXĐ: $\frac{5}{2}\geq x\geq \frac{3}{2}$
PT $\Leftrightarrow \sqrt{2x-3}+\sqrt{5-2x}=x^2-4x+6$
Ta thấy:
$\text{VP}=x^2-4x+6=(x-2)^2+2\geq 2$
Áp dụng BĐT Bunhiacopxky:
$\text{VT}^2\leq (2x-3+5-2x)(1+1)=4\Rightarrow \text{VT}\leq 2$
Do đó:
$\text{VT}\leq 2\leq \text{VP}$
Dấu "=" xảy ra khi \(\left\{\begin{matrix} \sqrt{2x-3}=\sqrt{5-2x}\\ (x-2)^2=0\end{matrix}\right.\) hay $x=2$
Vậy.......