ĐKXĐ: \(x>0;b\ne1\)
\(B=\left(\dfrac{1}{\sqrt{b}-1}-\dfrac{1}{\left(\sqrt{b}-1\right)\left(b+\sqrt{b}+1\right)}\right).\dfrac{3\left(\sqrt{b}-1\right)}{\sqrt{b}\left(\sqrt{b}+1\right)}\)
\(=\left(\dfrac{b+\sqrt{b}+1-1}{\left(\sqrt{b}-1\right)\left(b+\sqrt{b}+1\right)}\right).\dfrac{3\left(\sqrt{b}-1\right)}{\sqrt{b}\left(\sqrt{b}+1\right)}\)
\(=\dfrac{\sqrt{b}\left(\sqrt{b}+1\right).3\left(\sqrt{b}-1\right)}{\left(\sqrt{b}-1\right)\left(b+\sqrt{b}+1\right)\sqrt{b}\left(\sqrt{b}+1\right)}\)
\(=\dfrac{3}{b+\sqrt{b}+1}\)
b.
\(b+\sqrt{b}+1\ge0+0+1=1\Rightarrow B\le\dfrac{3}{1}=3\)
\(b+\sqrt{b}+1>0\Rightarrow\dfrac{3}{b+\sqrt{b}+1}>0\)
\(\Rightarrow0< B\le3\Rightarrow B=\left\{1;2;3\right\}\)
- Với \(B=1\Rightarrow\dfrac{3}{b+\sqrt{b}+1}=1\Rightarrow b+\sqrt{b}-2=0\Rightarrow b=1\) (ktm ĐKXĐ)
- Với \(B=2\Rightarrow\dfrac{3}{b+\sqrt{b}+1}=2\Rightarrow b+\sqrt{b}-1=0\)
\(\Rightarrow\sqrt{b}=\dfrac{\sqrt{5}-1}{2}\Rightarrow b=\dfrac{3-\sqrt{5}}{2}\)
- Với \(B=3\Rightarrow\dfrac{3}{b+\sqrt{b}+1}=3\Rightarrow b+\sqrt{b}=0\Rightarrow b=0\) (ktm ĐKXĐ)
Vậy \(b=\dfrac{3-\sqrt{5}}{2}\) thì B nguyên
