2.{1/9.10+1/10.11+1/11.12+.............+1/x.(x+1)=1/9
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25 tháng 2 2018
\(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}\)
\(=\frac{8-7}{7\cdot8}+\frac{9-8}{8\cdot9}+\frac{10-9}{9\cdot10}+\frac{11-10}{10\cdot11}+\frac{12-11}{11\cdot12}+\frac{13-12}{12\cdot13}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{7}-\frac{1}{13}=\frac{13-7}{7\cdot13}=\frac{6}{91}\)
19 tháng 11 2018
co \(\frac{1}{9\cdot10}=\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{10\cdot11}=\frac{1}{10}-\frac{1}{11}\)
............
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
nen \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\)
\(=\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}-...+\frac{1}{x}-\frac{1}{x+1}\)
=\(\frac{1}{9}-\frac{1}{x+1}\)
2 . ( \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\))
= 2 . ( \(\frac{1}{9}-\frac{1}{x+1}\)) = \(\frac{2}{9}-\frac{2}{x+1}\)
DN
0
KN
28 tháng 4 2019
\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{9}\div2\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=18-1\)
\(\Leftrightarrow x=17\)
KN
28 tháng 4 2019
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\Leftrightarrow\left|x\right|=\frac{20}{12}+\frac{9}{12}\)
\(\Leftrightarrow\left|x\right|=\frac{29}{12}\)
\(\Leftrightarrow x=\pm\frac{29}{12}\)
21 tháng 5 2020
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{13\cdot14}+\frac{1}{14\cdot15}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}\)
\(=\frac{1}{2}-\frac{1}{15}\)
\(=\frac{13}{30}\)
\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)
\(\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}.\frac{1}{2}\)
\(\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
\(\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}=\frac{1}{9}\)
=>x+1=9
=>x=8