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\(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}\)\(+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(=\frac{1}{4}+\frac{1}{3}\)
\(=\frac{7}{12}\)
Bài 2:
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{999\cdot1000}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)
=1-1/1000
=999/1000
Bài 1:
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+....+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{4033}{2017}\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{4033}{4034}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{4033}{4034}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4033}{4034}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{4033}{4034}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{4033}{4034}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4034}\)
\(\Rightarrow x+1=4034\)
\(\Rightarrow x=4034-1\)
\(\Rightarrow x=4033\)
\(A=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+..........+\frac{1}{8}.\frac{1}{9}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{8.9}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.......+\frac{1}{8}-\frac{1}{9}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
\(B=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+....+\frac{1}{110}=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{10.11}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)
\(\text{c,d cơ bản tự làm nha }\)
A=>1.1/2.3+1.1/3.4+1.1/4.5+1.1/5.6+1.11/6.7+.1/7.8+1.1/8.9
=>1/2.3+1/3.4+1/4.5+1/6.7+1/7.8+1/8.9
=>1/2-1/3-1/4-1/5-1/6-1/7-1/8-1/9
=>1/2-1/9=>9/18-2/18=>7/18
Vậy A= 7/18
\(A=\frac{61}{168}\)