\(P=\frac{x+3\sqrt{x-1}+1}{x+4\sqrt{x-1}+2}=\frac{\left(x-1\right)+3\sqrt{x-1}+2}{\left(x-1\right)+4\sqrt{x-1}+3}=\frac{y^2+3y+2}{y^2+4y+3}\) với \(y=\sqrt{x-1}\Rightarrow y\ge0\)
nên \(P=\frac{y+2}{y+3}=1-\frac{1}{y+3}\ge1-\frac{1}{3}=\frac{2}{3}\)
Dấu \(''=''\) xảy ra khi \(y=0\) hay \(x=1\)
Kết luận: ...