Lời giải:
ĐKXĐ: \(a\geq 0; a\neq 1\)
a) Ta có:
\(P=\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
\(=\frac{a+\sqrt{a}+1}{a+1}:\left(\frac{a+1}{(a+1)(\sqrt{a}-1)}-\frac{2\sqrt{a}}{a(\sqrt{a}-1)+(\sqrt{a}-1)}\right)\)
\(=\frac{a+\sqrt{a}+1}{a+1}: \frac{a+1-2\sqrt{a}}{(a+1)(\sqrt{a}-1)}=\frac{a+\sqrt{a}+1}{a+1}:\frac{(\sqrt{a}-1)^2}{(a+1)(\sqrt{a}-1)}\)
\(=\frac{a+\sqrt{a}+1}{a+1}: \frac{\sqrt{a}-1}{a+1}=\frac{a+\sqrt{a}+1}{\sqrt{a}-1}\)
b)
\(P>1\Leftrightarrow \frac{a+\sqrt{a}+1}{\sqrt{a}-1}>1(*)\)
TH1: \(a>1\)
\((*)\Leftrightarrow a+\sqrt{a}+1>\sqrt{a}-1\)
\(\Leftrightarrow a+2>0\) (luôn đúng)
TH2: \(0\leq a< 1\rightarrow \sqrt{a}-1< 0\)
\((*)\Leftrightarrow a+\sqrt{a}+1< \sqrt{a}-1\Leftrightarrow a+2< 0\) (vô lý vì $a\geq 0$)
Vậy để $P>1$ thì \(a>1\)
c)
\(a=19-8\sqrt{3}=19-2\sqrt{48}=16+3-2\sqrt{16.3}=(4-\sqrt{3})^2\)
\(\rightarrow \sqrt{a}=4-\sqrt{3}\)
Do đó \(P=\frac{a+\sqrt{a}+1}{\sqrt{a}-1}=\frac{19-8\sqrt{3}+4-\sqrt{3}+1}{4-\sqrt{3}-1}=\frac{24-9\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{-3+9(3-\sqrt{3})}{3-\sqrt{3}}=\frac{3}{\sqrt{3}-3}+9\)
