đkxđ: a > 0; a khác 1
a/ \(B=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}+1\right)^2}{a-1}-\dfrac{\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\dfrac{a-1}{\sqrt{a}}\right)\)
= (\(\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1}{a-1}+4\sqrt{a}\))\(\cdot\dfrac{a-1}{\sqrt{a}}\)
\(=\left(\dfrac{4\sqrt{a}}{a-1}+4\sqrt{a}\right)\cdot\dfrac{a-1}{\sqrt{a}}\)
\(=\dfrac{4\sqrt{a}+4\sqrt{a}\left(a-1\right)}{a-1}\cdot\dfrac{a-1}{\sqrt{a}}=\dfrac{4\sqrt{a}+4a\sqrt{a}-4\sqrt{a}}{\sqrt{a}}=\dfrac{4a\sqrt{a}}{\sqrt{a}}=4a\)
b/ a = \(\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{\sqrt{6}\left(2-\sqrt{6}\right)}{4-6}=\dfrac{2\sqrt{6}-6}{-2}=\dfrac{6-2\sqrt{6}}{2}\) thì :
\(B=4\cdot\dfrac{6-2\sqrt{6}}{2}=2\left(6-2\sqrt{6}\right)=12-4\sqrt{6}\)
c/ \(\sqrt{B}>B\Leftrightarrow\sqrt{4a}>4a\Leftrightarrow4a>16a^2\)
\(\Leftrightarrow4a-16a^2>0\)
\(\Leftrightarrow4a\left(1-4a\right)>0\)
\(\Leftrightarrow1-4a>0\)
\(\Leftrightarrow4a< 1\Leftrightarrow a< \dfrac{1}{4}\)
kết hợp với đkxđ ta có: \(0< a< \dfrac{1}{4}\)
Vậy..............
