Lời giải:
Sửa đề: \(B=\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}-\sqrt{5}}-....+\frac{1}{\sqrt{100}-\sqrt{101}}\)
Sử dụng công thức \(a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})\) với \(a,b>0\) ta có:
\(B=-\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{1}{\sqrt{4}-\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{4}}+....-\frac{1}{\sqrt{101}-\sqrt{100}}\)
\(=-\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}{\sqrt{3}-\sqrt{2}}+\frac{(\sqrt{4}-\sqrt{3})(\sqrt{4}+\sqrt{3})}{\sqrt{4}-\sqrt{3}}-\frac{(\sqrt{5}-\sqrt{4})(\sqrt{5}+\sqrt{4})}{\sqrt{5}-\sqrt{4}}+....-\frac{(\sqrt{101}-\sqrt{100})(\sqrt{101}+\sqrt{100})}{\sqrt{101}-\sqrt{100}}\)
\(=-(\sqrt{3}+\sqrt{2})+(\sqrt{4}+\sqrt{3})-(\sqrt{5}+\sqrt{4})+...-(\sqrt{101}+\sqrt{100})\)
\(=-\sqrt{101}-\sqrt{2}\)