Đáp án:
$ 1) a)\dfrac{99}{100}\\
2)
a)
x=305$
Giải thích các bước giải:
1)$a)\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\\
=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\
=1-\dfrac{1}{100}\\
=\dfrac{100}{100}-\dfrac{1}{100}\\
=\dfrac{99}{100}\\
2)
a)
\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x(x+3)}=\dfrac{101}{1540}\\
\Leftrightarrow 3.\left ( \dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x(x+3)}\right )=3.\dfrac{101}{1540}\\
\Leftrightarrow \dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x(x+3)}=\dfrac{303}{1540}\\
\Leftrightarrow \dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\\
\Leftrightarrow \dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\\
\Leftrightarrow \dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\\
\Leftrightarrow \dfrac{1}{x+3}=\dfrac{308}{1540}-\dfrac{303}{1540}\\
\Leftrightarrow \dfrac{1}{x+3}=\dfrac{5}{1540}\\
\Leftrightarrow x+3=\dfrac{1540}{5}=308\\
\Leftrightarrow x=305$
