Ta có:\(B=\dfrac{2}{2\sqrt{1}}+\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+...+\dfrac{2}{2\sqrt{48}}\)
\(B>\dfrac{2}{\sqrt{1}+\sqrt{2}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}+\dfrac{2}{\sqrt{3}+\sqrt{4}}+...+\dfrac{2}{\sqrt{48}+\sqrt{49}}\)
\(B>2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{49}-\sqrt{48}\right)\)
\(B>2\cdot\left(-1+\sqrt{49}\right)=12\)(đpcm)