Lời giải:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{149.150}\)
\(2A=1+\frac{1}{3.4}+\frac{1}{3.4}+...+\frac{1}{149.150}+\frac{1}{149.150}\)
\(>1+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{150.151}\)
\(=1+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{150}-\frac{1}{151}\)
\(=1+\frac{1}{3}-\frac{1}{151}>1\Rightarrow A> \frac{1}{2}\)