Lời giải:
\(299A=\frac{300-1}{1.300}+\frac{301-2}{2.301}+\frac{302-3}{3.302}+....+\frac{400-101}{101.400}\)

\(=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)

\(=(1+\frac{1}{2}+....+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(1)\)

Mặt khác:

$101B=\frac{102-1}{1.102}+\frac{103-2}{2.103}+...+\frac{400-299}{299.400}$

$=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}$

$=(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{299})-(\frac{1}{102}+\frac{1}{103}+....+\frac{1}{400})$

$=(1+\frac{1}{2}+...+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(2)$

Từ $(1);(2)\Rightarrow 299A=101B$

$\Rightarrow \frac{A}{B}=\frac{101}{299}$

\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)

\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{3012}+...+\frac{1}{101}-\frac{1}{400}\right)\)

\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{400}\right)\)

\(A=\frac{1}{299}.\frac{399}{400}\)

\(A=\frac{399}{119600}\)

\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)

\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}\right)\)

\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{400}\right)\)

\(B=\frac{1}{101}.\frac{399}{400}\)

\(B=\frac{399}{40400}\)

\(\Rightarrow\frac{A}{B}=\frac{399}{\frac{119600}{\frac{399}{40400}}}=\frac{101}{299}\)

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