Lời giải:
$A=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50})-2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50})-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25})$
$=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}=B$
Ta có đpcm.