Lời giải:
Ta có:
\(T=\frac{3+x}{x}+\frac{6-x}{3-x}=\frac{3}{x}+1+\frac{3}{3-x}+1\)
\(=3\left(\frac{1}{x}+\frac{1}{1-x}\right)+2=\frac{9}{x(3-x)}+2\)
Vì \(x\in [1,2]\Rightarrow x,3-x>0\)
Áp dụng BĐT Cauchy ngược dấu: \(x(3-x)\leq \left(\frac{x+3-x}{2}\right)^2=\frac{9}{4}\)
\(\Rightarrow T\geq \frac{9}{\frac{9}{4}}+2=6\) hay \(T_{\min}=6\)
Dấu bằng xảy ra khi \(x=3-x\Leftrightarrow x=\frac{3}{2}\)
------------
Mặt khác: \(1\leq x\leq 2\Rightarrow (x-1)(x-2)\leq 0\)
\(\Leftrightarrow 3x-x^2\geq 2\Leftrightarrow x(3-x)\geq 2\)
\(\Rightarrow T\leq \frac{9}{2}+2=\frac{13}{2}\)
Vậy \(T_{\max}=\frac{13}{2}\Leftrightarrow \text{x=1 or x=2} \)