Áp dụng bất đẳng thức AM - GM, ta có:
\(S=\dfrac{1}{\left(x-1\right)^2}+\dfrac{1}{\left(2-x\right)^2}+\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\ge3\sqrt[3]{\dfrac{1}{\left(x-1\right)^2}\times\dfrac{1}{\left(2-x\right)^2}\times\dfrac{1}{\left(x-1\right)\left(2-x\right)}}\)
\(=\dfrac{3}{\left(x-1\right)\left(x-2\right)}=\dfrac{3}{-x^2+3x-2}\)
Vì \(-x^2+3x-2=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
nên \(S\ge\dfrac{3}{\dfrac{1}{4}}=12\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{\left(x-1\right)^2}=\dfrac{1}{\left(2-x\right)^2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\\x-\dfrac{3}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3}{2}\left(\text{ nhận }\right)\)
Vậy \(Min_S=12\Leftrightarrow x=\dfrac{3}{2}\)