a) ĐKXĐ: x ≠ 1; x ≥ 0
\(P=\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)-\left(x+2\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{x\sqrt{x}+2x-\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
b) \(Q=\)\(\frac{2}{P}+\sqrt{x}=\frac{2\left(x+\sqrt{x}+1\right)}{\sqrt{x}+2}+\sqrt{x}=\frac{3x+4\sqrt{x}+2}{\sqrt{x}+2}\)
\(Q=\frac{\left(3x+6\sqrt{x}\right)-\left(2\sqrt{x}+4\right)+6}{\sqrt{x}+2}=3\sqrt{x}-2+\frac{6}{\sqrt{x}+2}\)
\(Q=3\left(\sqrt{x}+2\right)-8+\frac{6}{\sqrt{x}+2}\ge2\sqrt{3.\left(\sqrt{x}+2\right).\frac{6}{\sqrt{x}+2}}-8=6\sqrt{2}-8\)
Dấu "=" xảy ra khi \(x=6\pm4\sqrt{2}\)
Vậy GTNN của Biểu thức Q là \(6\sqrt{2}-8\)