Bài 1:Tính: a,1/2+1/2.3+1/3.4+1/4.5+…+1/99/100 Bài 2:Tìm x: a,1/5.8+1/8.11+1/11.14+…1/x.(x+3)=101/1540

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1. Đáp án:

   $ 1) a)\dfrac{99}{100}\\
   2)
   a)
   x=305$

   Giải thích các bước giải:

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

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