a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\end{matrix}\right.\)
\(P=\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}-\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)+\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)-ab+1}{ab-1}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)-\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)+ab-1}{ab-1}\)
\(=\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}-ab+1}{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-\sqrt{ab}-a\sqrt{b}-\sqrt{a}+ab-1}\)
\(=\dfrac{2a\sqrt{b}+2\sqrt{ab}}{-2\sqrt{a}-2}\)
\(=\dfrac{-\sqrt{ab}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=-\sqrt{ab}\)
b: \(b=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{\left(\sqrt{3}-1\right)^2}{2}=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
Khi \(a=2+\sqrt{3};b=2-\sqrt{3}\) thì \(a\cdot b=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=4-3=1\)
=>\(P=-\sqrt{ab}=-1\)
