\(T=9x+\dfrac{1}{x-1}=9\left(x-1\right)+\dfrac{1}{x-1}+9\)
Áp dụng BĐT Cauchy cho 2 số dương:
\(9\left(x-1\right)+\dfrac{1}{x-1}\ge2\sqrt{9\left(x-1\right).\dfrac{1}{x-1}}=2.3=6\)
\(\Rightarrow T=9\left(x-1\right)+\dfrac{1}{x-1}\ge6+9=15\)
\(minT=15\Leftrightarrow9\left(x-1\right)=\dfrac{1}{x-1}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{9}\Leftrightarrow x=\dfrac{4}{3}\)(do \(x>1\))