ĐKXĐ:\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a>0\\b>0\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\b< 0\end{matrix}\right.\end{matrix}\right.\) và \(a\ne b\)
a)\(N=\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\)
\(=\frac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\frac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\frac{a+b}{\sqrt{ab}}\)
\(=\frac{a\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)+b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(b-a\right)}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}-\sqrt{a}\right)}\)
\(=\frac{a\sqrt{ab}-a^2+b^2+b\sqrt{ab}+b^2-a^2}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}-\sqrt{a}\right)}\)
\(=\frac{\sqrt{ab}\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}\)
\(=\frac{a+b}{b-a}\)
b)\(a=\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(b=\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
Khi đó, \(N=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{3}-1-\sqrt{3}-1}=\frac{2\sqrt{3}}{-2}=-\sqrt{3}\)
c)\(\frac{a}{b}=\frac{a+1}{b+5}\Leftrightarrow ab+5a=ab+b\Leftrightarrow5a=b\)
Thay vào N ta được:
\(N=\frac{a+5a}{5a-a}=\frac{6a}{4a}=\frac{3}{2}\)(luôn không đổi)
