Đáp án:
\(E = - \dfrac{{97}}{{99}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{{3 - 2}}{6} = \dfrac{1}{6}\\
B = \dfrac{1}{3} - \dfrac{1}{4} = \dfrac{{4 - 3}}{{12}} = \dfrac{1}{{12}}\\
C = \dfrac{1}{4} - \dfrac{1}{5} = \dfrac{{5 - 4}}{{20}} = \dfrac{1}{{20}}\\
b.A + B = \dfrac{1}{6} + \dfrac{1}{{12}} = \dfrac{{2 + 1}}{{12}} = \dfrac{3}{{12}} = \dfrac{1}{4}\\
A + B + C = \dfrac{1}{4} + \dfrac{1}{{20}} = \dfrac{{5 + 1}}{{20}} = \dfrac{6}{{20}} = \dfrac{3}{{10}}\\
c.D = \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{19.20}}\\
= \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{{19}} - \dfrac{1}{{20}}\\
= \dfrac{1}{2} - \dfrac{1}{{20}} = \dfrac{{10 - 1}}{{20}} = \dfrac{9}{{20}}\\
E = \dfrac{1}{{99}} - \left( {\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + ... + \dfrac{1}{{98.99}}} \right)\\
= \dfrac{1}{{99}} - \left( {1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{{98}} - \dfrac{1}{{99}}} \right)\\
= \dfrac{1}{{99}} - \left( {1 - \dfrac{1}{{99}}} \right)\\
= \dfrac{1}{{99}} - 1 + \dfrac{1}{{99}}\\
= \dfrac{2}{{99}} - 1 = \dfrac{{2 - 99}}{{99}} = - \dfrac{{97}}{{99}}
\end{array}\)
