$\displaystyle \begin{array}{{>{\displaystyle}l}} B=\frac{1+\sqrt{b}}{1-\sqrt{b}} -\frac{1-\sqrt{b}}{1+\sqrt{b}} +\frac{4b}{1-b} \ \\ DKXD\ :\ \begin{cases} b\geqslant 0 & \\ b\#1 & \end{cases} \ \\ B=\frac{\left( 1+\sqrt{b}\right)^{2} -\left( 1-\sqrt{b}\right)^{2} +4b}{1-b} \ \\ B=\frac{1+2\sqrt{b} +b-1+2\sqrt{b} -b+4b}{1-b} \ \\ B=\frac{4\sqrt{b} +4b}{1-b} =\frac{4\sqrt{b}\left(\sqrt{b} +1\right)}{\left( 1-\sqrt{b}\right)\left( 1+\sqrt{b}\right)} =\frac{4\sqrt{b}}{1-\sqrt{b}} \ \\ b) \ B >-1\ \\ \rightarrow \frac{4\sqrt{b}}{1-\sqrt{b}} >-1\ \rightarrow \frac{4\sqrt{b}}{1-\sqrt{b}} +1 >0\ \\ \frac{4\sqrt{b} +1-\sqrt{b}}{1-\sqrt{b}} >0\ \\ \frac{3\sqrt{b} +1}{1-\sqrt{b}} \ >0\ \\ Nhận\ thấy\ 3\sqrt{b} +1\ >0\ với\ mọi\ b\ \\ do\ đó\ để\ B\ +1 >0\ \\ \rightarrow 1-\sqrt{b} >0\ \rightarrow \sqrt{b} < 1\ hay\ b< 1\ \\ Vậy\ 0\leqslant b< 1\ thì\ B\ nguyên\ \end{array}$
