Đặt \(\sqrt{x-1}=a\ge0\Rightarrow x=a^2+1\)
\(P=\dfrac{a^2+3a+2}{a^2+4a+3}=\dfrac{\left(a+1\right)\left(a+2\right)}{\left(a+1\right)\left(a+3\right)}=\dfrac{a+2}{a+3}=1-\dfrac{1}{a+3}\)
Do \(\dfrac{1}{a+3}>0\Rightarrow P_{min}\) khi \(\dfrac{1}{a+3}\) lớn nhất \(\Rightarrow a+3\) nhỏ nhất
Mà \(a+3\ge3\Rightarrow P_{min}=1-\dfrac{1}{3}=\dfrac{2}{3}\) khi \(a=0\Leftrightarrow x=1\)