Lời giải:
\(15A=\frac{17-2}{2.17}+\frac{18-3}{3.18}+\frac{19-4}{4.19}+...+\frac{2005-1990}{1990.2005}\)
\(=\frac{1}{2}-\frac{1}{17}+\frac{1}{3}-\frac{1}{18}+\frac{1}{4}-\frac{1}{19}+...+\frac{1}{1990}-\frac{1}{2005}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1990}\right)-\left(\frac{1}{17}+\frac{1}{18}+..+\frac{1}{2005}\right)\)
\(=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}-\left(\frac{1}{1991}+\frac{1}{1992}+...+\frac{1}{2005}\right)\)
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\(1989B=\frac{1991-2}{2.1991}+\frac{1992-3}{3.1992}+\frac{1993-4}{4.1993}+...+\frac{2005-16}{16.2005}\)
\(=\frac{1}{2}-\frac{1}{1991}+\frac{1}{3}-\frac{1}{1992}+\frac{1}{4}-\frac{1}{1993}+...+\frac{1}{16}-\frac{1}{2005}\)
\(=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}-\left(\frac{1}{1991}+\frac{1}{1992}+...+\frac{1}{2005}\right)\)
Do đó:
\(\frac{15A}{1989B}=1\Leftrightarrow \frac{A}{B}=\frac{1989}{15}\)