Bài 2:
a) Thay \(x=25-4\sqrt{6}\) vào A, ta được:
\(A=\dfrac{25-4\sqrt{6}+12}{2\sqrt{6}-1-1}\)
\(=\dfrac{37-4\sqrt{6}}{2\sqrt{6}-2}\)
\(=\dfrac{\left(37-4\sqrt{6}\right)\left(2\sqrt{6}+2\right)}{20}\)
\(=\dfrac{74\sqrt{6}+74-48-4\sqrt{6}}{20}\)
\(=\dfrac{70\sqrt{6}+26}{20}=\dfrac{35\sqrt{6}+13}{10}\)
b) Ta có: \(B=\left(\dfrac{3}{x-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}+1}\)
\(=\left(\dfrac{3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
d) Để B nguyên thì \(\sqrt{x}+2⋮\sqrt{x}-1\)
\(\Leftrightarrow3⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2;4\right\}\)
hay \(x\in\left\{0;4;16\right\}\)
