a, Đk: x ≥ 0, x ≠ 1
\(P=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)
\(=\left[\frac{x+2}{\left(\sqrt{x}\right)^3-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right].\frac{2}{\sqrt{x}-1}\)
\(=\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2.\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2}{x+\sqrt{x}+1}\)
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b, Để P = \(\frac{2}{7}\)
⇔ \(\frac{2}{x+\sqrt{x}+1}=\frac{2}{7}\)
⇔ \(x+\sqrt{x}+1=7\)
⇔ \(x+\sqrt{x}-6=0\)
⇔ \(x-2\sqrt{x}+3\sqrt{x}-6=0\)
⇔ \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)=0\)
Vì x ≥ 0 ⇒ \(\sqrt{x}+3>0\)
nên \(\sqrt{x}-2=0\)
⇔ \(\sqrt{x}=2\)
⇔ x = 4
Vậy x = 4 thì P = \(\frac{2}{7}\)
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c, Vì x ≥ 0 ⇒ \(x+\sqrt{x}+1\ge1\)
⇔ \(0< \frac{2}{x+\sqrt{x}+1}\le2\)
⇔ 0 < P ≤ 2
⇔ P - 2 ≤ 0
⇔ P(P - 2) ≤ 0
⇔ P2 -2P ≤ 0
⇔ P2 ≤ 2P
Dấu "=" xảy ra khi P = 3 ⇒ x = 0
Vậy P2 ≤ 2P
