Lời giải:
ĐKXĐ: \(b> 0; b\neq 1; b\neq 4\)
1.
\(B=\left(\frac{2+\sqrt{b}}{1-\sqrt{b}}-\frac{1+\sqrt{b}}{2-\sqrt{b}}\right):\left(\frac{1}{1-\sqrt{b}}+\frac{1}{\sqrt{b}}\right)\)
\(=\frac{(2+\sqrt{b})(2-\sqrt{b})-(1+\sqrt{b})(1-\sqrt{b})}{(1-\sqrt{b})(2-\sqrt{b})}:\frac{\sqrt{b}+1-\sqrt{b}}{(1-\sqrt{b}).\sqrt{b}}\)
\(=\frac{4-b-(1-b)}{(1-\sqrt{b})(2-\sqrt{b})}:\frac{1}{(1-\sqrt{b}).\sqrt{b}}\)
\(=\frac{3}{(1-\sqrt{b})(2-\sqrt{b})}.(1-\sqrt{b}).\sqrt{b}=\frac{3\sqrt{b}}{2-\sqrt{b}}\)
2.
\(B\sqrt{b}-1=0\Leftrightarrow \frac{3b}{2-\sqrt{b}}-1=0\)
\(\Leftrightarrow 3b-(2-\sqrt{b})=0\)
\(\Leftrightarrow 3b+\sqrt{b}-2=0\)
\(\Leftrightarrow (3\sqrt{b}-2)(\sqrt{b}+1)=0\)
\(\Rightarrow 3\sqrt{b}-2=0\Rightarrow b=\frac{4}{9}\) (thỏa mãn)
Vậy.........
