BT
2
K
Khách
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5 tháng 5 2018
\(\frac{8^2}{7.9}.\frac{9^2}{8.10}...\frac{14^2}{13.15}\)
\(\frac{8.8}{7.9}.\frac{9.9}{8.10}...\frac{14.14}{13.15}\)
\(\frac{8.9...14}{7.8...13}.\frac{8.9...14}{9.10...15}\)
\(\frac{14}{7}.\frac{8}{15}\)
\(2.\frac{8}{15}\)
\(\frac{16}{15}\)
5 tháng 5 2018
(8.9.10.11.12.13.14)(8.9.10.11.12.13.14)/7.8.9.10.11.12.13).(9.10.11.12.13.14.15)
=14.8/7.15
=16/15
k cho mình nhá
QB
0
AH
0
OY
22 tháng 7 2021
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
22 tháng 7 2021
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
DX
6 tháng 8 2021
Ta có \(A=\dfrac{2}{1.3}-\dfrac{2}{2.4}+\dfrac{2}{3.5}-\dfrac{2}{4.6}+\dfrac{2}{5.7}-\dfrac{2}{6.8}+\dfrac{2}{7.9}-\dfrac{2}{8.10}+\dfrac{2}{9.11}-\dfrac{2}{10.12}\)
\(\Rightarrow A=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)-\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+\dfrac{2}{10.12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
\(\Rightarrow A=1-\dfrac{1}{11}-\dfrac{1}{2}+\dfrac{1}{12}\)
\(\Rightarrow A=\dfrac{9}{22}+\dfrac{1}{12}\)
\(\Rightarrow A=\dfrac{65}{132}\)
Mà \(\dfrac{65}{132}< 1\) \(\Rightarrow A< 1\)
Vậy \(A< 1\)
T
1
18 tháng 5 2018
\(\frac{x}{3.5}+\frac{x}{5.7}+\frac{x}{7.9}+...+\frac{x}{13.15}=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{13.15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\frac{4}{15}=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}=\frac{4}{45}:\frac{4}{15}\)
\(\Leftrightarrow\frac{x}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}.2\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
_Chúc bạn học tốt_
PT
1
CM
14 tháng 8 2019
Ta có: 102+112+122 = 100 + 121 + 144 = 365
132+142 = 169 + 196 = 365
Vậy 102+112+122 = 132+142
Ta có:\(\frac{8^2}{7.9}.\frac{9^2}{8.10}.\frac{10^2}{9.11}...\frac{14^2}{13.15}=\frac{8^2.9^2.....14^2}{7.9.8.10.9.11....13.15}\)
\(=\)\(\frac{\left(8.9.10...14\right)\left(8.9.10...14\right)}{\left(7.8.9...13\right).\left(9.10.11...15\right)}\)
\(=\frac{14.8}{7.15}=\frac{2.7.8}{7.15}=\frac{2.8}{15}=\frac{16}{15}\)
\(\frac{8^2}{7.9}.\frac{9^2}{8.10}.\frac{10^2}{9.11}.\frac{11^2}{10.12}.\frac{12^2}{11.13}.\frac{13^2}{12.14}.\frac{14^2}{13.15}\)
\(\frac{8^2.9^2.10^2.11^2.12^2.13^2.14^2}{7.9.8.10.9.11.10.12.11.13.12.14.13.15}\)
\(\frac{8.9.10.11.12.13.14}{7.9.10.11.12.13.15}=\frac{8.14}{7.15}=\frac{112}{105}=\frac{16}{15}\)
Học tốt@_@