Ta có: \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).....\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}=\frac{3.4.5..100}{2.3.4...99}=\frac{100}{2}=50\)
Vậy \(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)......\left(\frac{1}{99}+1\right)=50\)
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