Đáp án:
$7)
a) \dfrac{2019}{2020}\\
b) \dfrac{102}{309}\\
c)
\dfrac{-1999}{2000}\\
d)
\dfrac{-50}{101}$
Giải thích các bước giải:
$7)
a) \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2019.2020}\\
=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\\
=1-\dfrac{1}{2020}\\
=\dfrac{2020}{2020}-\dfrac{1}{2020}\\
=\dfrac{2019}{2020}\\
b) \dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{101.103}\\
=\dfrac{1}{3}.\left ( \dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{101.103} \right )\\
=\dfrac{1}{3}.\left ( \dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{101}-\dfrac{1}{103} \right )\\
=\dfrac{1}{3}.\left ( 1-\dfrac{1}{103} \right )\\
=\dfrac{1}{3}.\left ( \dfrac{103}{103}-\dfrac{1}{103} \right )\\
=\dfrac{1}{3}.\dfrac{102}{103}\\
=\dfrac{102}{309}\\
c)
-\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\\
=-\left ( \dfrac{1}{2000.1999}+\dfrac{1}{1999.1998}+\dfrac{1}{1998.1997}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1} \right )\\
=-\left ( \dfrac{1}{2.1}+\dfrac{1}{3.2}+...+\dfrac{1}{1998.1997}+\dfrac{1}{1999.1998}+\dfrac{1}{2000.1999} \right )\\
=-\left ( 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1997}-\dfrac{1}{1998}+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}-\dfrac{1}{2000} \right )\\
=-\left ( 1-\dfrac{1}{2000} \right )\\
=-1+\dfrac{1}{2000}\\
=\dfrac{-2000}{2000}+\dfrac{1}{2000}\\
=\dfrac{-1999}{2000}\\
d)
-\dfrac{1}{3}+\dfrac{-1}{15}+\dfrac{-1}{35}+\dfrac{-1}{63}+...+\dfrac{-1}{9999}\\
=-\left ( \dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{999}\right )\\
=-\left ( \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}\right )\\
=-\dfrac{1}{2}.\left ( \dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\right )\\
=-\dfrac{1}{2}.\left ( 1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\right )\\
=\dfrac{-1}{2}.\left ( 1-\dfrac{1}{101} \right )\\
=\dfrac{-1}{2}.\left ( \dfrac{101}{101}-\dfrac{1}{101} \right )\\
=\dfrac{-1}{2}.\dfrac{100}{101}\\
=\dfrac{-50}{101}$