Đáp án:
\(\begin{array}{l}
a.\frac{{101}}{{102}}\\
b.\frac{{50}}{{51}}\\
c.\frac{7}{{36}}\\
d.\frac{{41}}{{254}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\frac{1}{{1.2}} + \frac{1}{{2.3}} + ... + \frac{1}{{101.102}}\\
= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{{101}} - \frac{1}{{102}}\\
= 1 - \frac{1}{{102}} = \frac{{101}}{{102}}\\
b.\frac{2}{{1.3}} + \frac{2}{{3.5}} + ... + \frac{2}{{49.51}}\\
= 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + ... + \frac{1}{{49}} - \frac{1}{{51}}\\
= 1 - \frac{1}{{51}} = \frac{{50}}{{51}}\\
c.\frac{3}{{3.8}} + \frac{3}{{8.13}} + ... + \frac{3}{{103.108}}\\
= 3.\frac{1}{5}.\left( {\frac{5}{{3.8}} + \frac{5}{{8.13}} + ... + \frac{5}{{103.108}}} \right)\\
= \frac{3}{5}\left( {\frac{1}{3} - \frac{1}{8} + \frac{1}{8} - \frac{1}{{13}} + ... + \frac{1}{{103}} - \frac{1}{{108}}} \right)\\
= \frac{3}{5}.\left( {\frac{1}{3} - \frac{1}{{108}}} \right) = \frac{7}{{36}}\\
d.\frac{2}{{4.7}} + \frac{2}{{7.10}} + ... + \frac{2}{{124.127}}\\
= 2.\frac{1}{3}.\left( {\frac{3}{{4.7}} + \frac{3}{{7.10}} + ... + \frac{3}{{124.127}}} \right)\\
= \frac{2}{3}.\left( {\frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + ... + \frac{1}{{124}} - \frac{1}{{127}}} \right)\\
= \frac{2}{3}.\left( {\frac{1}{4} - \frac{1}{{127}}} \right) = \frac{{41}}{{254}}
\end{array}\)
