\(VT=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{48}+\sqrt{49}}=\dfrac{\sqrt{2}-1}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}+\dfrac{\sqrt{4}-\sqrt{3}}{1}+...+\dfrac{\sqrt{49}-\sqrt{48}}{1}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{49}-\sqrt{48}=\sqrt{49}-1=7-1=6=VP\)Vậy , đẳng thức được CM