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TC
2
6 tháng 3 2022
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{15}-\dfrac{1}{18}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{18}\right)=\dfrac{1}{3}\cdot\dfrac{5}{18}=\dfrac{5}{54}\)
NH
1
15 tháng 3 2016
\(\frac{4}{3.6}+\frac{4}{6.9}+...+\frac{4}{12.15}\)
\(=\frac{4\left(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{12.15}\right)}{3}\)
\(=\frac{4\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{12}-\frac{1}{15}\right)}{3}\)
\(=\frac{4\left(\frac{1}{3}-\frac{1}{15}\right)}{3}\)
\(=\frac{\frac{16}{15}}{3}=\frac{48}{15}\)
DN
2
WH
7 tháng 5 2018
Đặt \(A=\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{96\cdot99}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{99}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{99}\)
\(\Rightarrow A=\frac{32}{99}\)
13 tháng 5 2018
\(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{96.99}\)
\(=\frac{3}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{99}\right)\)
\(=1.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=1.\frac{32}{99}\)
\(=\frac{32}{99}\)
NT
1
29 tháng 5 2020
\(S=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{219.222}\)
\(\Rightarrow3S=\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{219.222}\)
\(=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{219}-\frac{1}{222}\)
\(=\frac{1}{3}-\frac{1}{222}< \frac{1}{3}\)
\(\Rightarrow S< \frac{1}{9}< 1\)
\(\Rightarrow S< 1\left(đpcm\right)\)
VT
1
23 tháng 2 2022
\(N=\dfrac{2}{3.6}+\dfrac{2}{6.9}+...+\dfrac{2}{2019.2022}\)
\(\Rightarrow N=2\left(\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{2019.2022}\right)\)
\(\Rightarrow N=\dfrac{2}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{2019.2022}\right)\)
\(\Rightarrow N=\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{2019}-\dfrac{1}{2022}\right)\)
\(\Rightarrow N=\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2022}\right)\)
\(\Rightarrow N=\dfrac{2}{3}.\dfrac{673}{2022}\\ \Rightarrow N=\dfrac{673}{3033}\)
TT
2
TT
1
Sửa đề: \(\dfrac{5}{3\cdot6}+\dfrac{5}{6\cdot9}+\dfrac{5}{9\cdot12}+\dfrac{5}{12\cdot15}\)
\(=\dfrac{5}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+\dfrac{3}{12\cdot15}\right)\)
\(=\dfrac{5}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{15}\right)\)
\(=\dfrac{5}{3}\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{5}{3}\cdot\dfrac{4}{15}=\dfrac{20}{45}=\dfrac{4}{9}\)