2/10.12+2/12.14-2/12.14+....+2/48.50
tra loi cho minh nhe minh tich cho
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S=\(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+.....+\frac{2}{98.100}\)
S=\(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+........+\frac{1}{98}-\frac{1}{100}\)
S=\(\frac{1}{10}-\frac{1}{100}\)
S=\(\frac{9}{100}\)<\(\frac{1}{10}\)
\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+...+\dfrac{2}{98\cdot100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{100}=\dfrac{9}{50}=0,18\)
Vậy \(S>\dfrac{1}{10}\)
\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+\dfrac{2}{14\cdot16}+...+\dfrac{2}{98\cdot100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{100}\)
\(S=\dfrac{20}{100}-\dfrac{2}{100}\)
\(S=\dfrac{18}{100}=\dfrac{9}{50}=0,18\)
\(\dfrac{1}{10}=0,1\), mà \(0,1< 0,18\)
\(\Rightarrow S>\dfrac{1}{10}\left(đpcm\right)\)
a) A = 3/10.12 + 3/12.14 + ... + 3/998.1000
2/3.A = 2/10.12 + 2/12.14 + ... + 2/998.1000
2/3.A = 1/10 - 1/12 + 1/12 - 1/14 + ... + 1/998 - 1/1000
2/3.A = 1/10 - 1/1000
2/3.A = 99/1000
A = 99/1000 : 2/3
A = 99/1000 . 3/2
A = 297/2000
b) B = 2/1.4 + 2/4.7 + 2/7.10 + ... + 2/22.25
3/2.B = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/22.25
3/2.B = 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/22 - 1/25
3/2.B = 1 - 1/25
3/2.B = 24/25
B = 24/25 : 3/2
B = 24/25 . 2/3
B = 16/25
Ủng hộ mk nha ^_-
a) Ta có: \(A=\frac{3}{10.12}+\frac{3}{12.14}+....+\frac{3}{998.1000}.\)
\(\Rightarrow\frac{2}{3}A=\frac{1}{10.12}+\frac{1}{12.14}+...+\frac{1}{998.1000}\)
\(\Rightarrow\frac{2}{3}A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{998}-\frac{1}{1000}\)
\(\Rightarrow\frac{2}{3}A=\frac{1}{10}-\frac{1}{1000}=\frac{99}{1000}\)
\(\Rightarrow A=\frac{99}{1000}:\frac{2}{3}=\frac{297}{2000}\)
Đặt \(A=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{48\cdot50}\)
\(A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
\(A=\frac{1}{10}-\frac{1}{50}=\frac{5}{50}-\frac{1}{50}=\frac{4}{50}=\frac{2}{25}\)
Vậy \(A=\frac{2}{25}\)
= \(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
= \(\frac{1}{10}-\frac{1}{50}\)= \(\frac{2}{25}\)
Đặt tổng trên là A ta có
\(2A=\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{48.52}\)
\(2A=\frac{12-10}{10.12}+\frac{14-12}{12.14}+\frac{16-14}{14.16}+...+\frac{50-48}{48.50}\)
\(2A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{48}-\frac{1}{50}=\frac{1}{10}-\frac{1}{50}=\frac{2}{25}\)
\(\Rightarrow A=\frac{2A}{2}=\frac{1}{25}\)
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@_@