a) Điều kiện xác định: \(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}\ne0\\\sqrt{a}-2\ne0\\\sqrt{a}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>0\\\sqrt{a}\ne2\\\sqrt{a}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>0\\a\ne4\\a\ne1\end{matrix}\right.\)
b) \(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
c) \(B>\dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>\dfrac{1}{6}\Leftrightarrow6\left(\sqrt{a}-2\right)>3\sqrt{a}\)
\(\Leftrightarrow6\sqrt{a}-12>3\sqrt{a}\Leftrightarrow3\sqrt{a}>12\Leftrightarrow\sqrt{a}>4\Leftrightarrow a>16\) (TM)
Vậy \(a>16\) thì \(B>\dfrac{1}{6}\)
d) \(B=\dfrac{\sqrt{a}-2}{3\sqrt{a}}=\dfrac{1}{3}-\dfrac{2}{3\sqrt{a}}\)
Do \(a\) nguyên thỏa mãn \(a>0,a\ne1,a\ne4\)
\(\Rightarrow a\) nguyên, \(a\ge2\)
\(\Rightarrow\sqrt{a}\ge\sqrt{2}\Rightarrow B\ge\dfrac{1}{3}-\dfrac{2}{3\sqrt{2}}=\dfrac{1-\sqrt{2}}{3}\)
Dấu = xảy ra \(\Leftrightarrow a=2\)
Vậy khi \(a=2\) thì \(B_{min}=\dfrac{1-\sqrt{2}}{3}\)