Đáp án:
$\frac{e}{f}=\frac{1}{10}$
Giải thích các bước giải:
$e=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\\
=\frac{1}{100}.\left ( \frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110} \right )\\
=\frac{1}{100}.\left ( 1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )\\
=\frac{1}{100}.\left [\left ( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10} \right)-\left ( \frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right ) \right ]\\
f=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\\
=\frac{1}{10}.\left ( \frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110} \right )\\
=\frac{1}{10}.\left ( 1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )\\
=\frac{1}{10}.\left [ \left ( 1+\frac{1}{2}+...+\frac{1}{100} \right )-\left ( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{110} \right ) \right ]\\
=\frac{1}{10}.\left [\left (1+\frac{1}{2}+...+\frac{1}{10} \right )- \left ( \frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right ) \right ]\\
=10e\\
\Rightarrow \frac{e}{f}=\frac{e}{10e}=\frac{1}{10}$
