\(ĐKXĐ:a\ne1;a\ge0;\)
\(B=\left(1+\dfrac{\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)-\left(a+1\right)}\right]\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\left[\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}:\dfrac{\sqrt{a}-1}{a+1}\)
\(B=\dfrac{a+1+\sqrt{a}}{a+1}.\dfrac{a+1}{\sqrt{a}-1}\)
\(B=\dfrac{a+1+\sqrt{a}}{\sqrt{a}-1}\)
b.
\(B=\dfrac{a+1+\sqrt{a}}{\sqrt{a}-1}>1\)
\(\Rightarrow B=\dfrac{a+1+\sqrt{a}}{\sqrt{a}-1}-1>0\)
\(\Rightarrow B=\dfrac{a+1+\sqrt{a}-\sqrt{a}+1}{\sqrt{a}-1}>0\)
\(\Rightarrow B=\dfrac{a+2}{\sqrt{a}-1}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+2>0\\\sqrt{a}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}a+2< 0\\\sqrt{a}-1< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a>1\\a< -2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow a>1\)
c.
\(a=2007-2\sqrt{2006}=2006-2\sqrt{2006}+1=\left(\sqrt{2006}-1\right)^2\)Thay \(a\) vào ta được:
\(B=\dfrac{2007-2\sqrt{2006}+1+\sqrt{\left(\sqrt{2006}-1\right)^2}}{\sqrt{\left(\sqrt{2006}-1\right)^2}-1}\)
\(B=\dfrac{2007-2\sqrt{2006}+1+\sqrt{2006}-1}{\sqrt{2006}-1-1}\)
\(B=\dfrac{2007-\sqrt{2006}}{\sqrt{2006}-2}\)
\(B=\dfrac{\left(2007-\sqrt{2006}\right)\left(\sqrt{2006}+2\right)}{\left(\sqrt{2006}-2\right)\left(\sqrt{2006}+2\right)}\)
\(B=\dfrac{2007\sqrt{2006}+4014-2006-2\sqrt{2006}}{2006-4}\)
\(B=\dfrac{2005\sqrt{2006}+2008}{2002}\)