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S = \(\dfrac{1}{10.12}\) + \(\dfrac{1}{12.14}\) + .....+ \(\dfrac{1}{998.1000}\)
S = \(\dfrac{1}{2}\).( \(\dfrac{2}{10.12}\) + \(\dfrac{2}{12.14}\)+.....+ \(\dfrac{2}{998.1000}\))
S = \(\dfrac{1}{2}\).( \(\dfrac{1}{10}\)- \(\dfrac{1}{12}\)+ \(\dfrac{1}{12}\) - \(\dfrac{1}{14}\)+...+\(\dfrac{1}{998}\)- \(\dfrac{1}{1000}\))
S = \(\dfrac{1}{2}\). ( \(\dfrac{1}{10}\) - \(\dfrac{1}{1000}\))
S = \(\dfrac{1}{2}\).\(\dfrac{99}{1000}\)
S = \(\dfrac{99}{2000}\)
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 \(A=\frac{2}{25}\)
= \(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
= \(\frac{1}{10}-\frac{1}{50}\)= \(\frac{2}{25}\)
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 tổng trên là A ta có
\(2A=\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{48.52}\)
\(2A=\frac{12-10}{10.12}+\frac{14-12}{12.14}+\frac{16-14}{14.16}+...+\frac{50-48}{48.50}\)
\(2A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{48}-\frac{1}{50}=\frac{1}{10}-\frac{1}{50}=\frac{2}{25}\)
\(\Rightarrow A=\frac{2A}{2}=\frac{1}{25}\)
\(\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)
=\(3.\left(\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)
=\(\frac{3}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\right)\)
=\(\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)
=\(\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)
=\(\frac{3}{2}.\left(\frac{7}{28}-\frac{2}{28}\right)\)
=\(\frac{3}{2}.\frac{5}{28}=\frac{15}{56}\)
\(\sqrt[]{\frac{ }{ }\frac{ }{ }\hept{\begin{cases}\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}^2}\)
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)=\frac{5}{2}.\frac{37}{150}=\frac{37}{60}\)
\(A=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}....\frac{999^2}{998.1000}\)
\(A=\frac{2^2.3^2....999^2}{1.3.2.4.998.100}=\frac{\left(2.3.....999\right)\left(2.3....999\right)}{\left(1.2....998\right)\left(3.4....1000\right)}\)
\(A=999\cdot\frac{1}{500}=\frac{999}{500}\)( khúc này mk làm tắt, bn bỏ dấu ở trên rồi bỏ từng tử)
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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=\(\frac{9}{100}\)<\(\frac{1}{10}\)
2/5 . A = 2/10.12 + 2/12.14 + ....... + 2/998.1000
= 1/10 - 1/12 + 1/12 - 1/14 + ..... + 1/998 - 1/1000
= 1/10 - 1/1000 = 99/1000
=> A = 99/1000 : 2/5 = 99/400
Tk mk nha
Mình làm tắt cũng dc nhé có gì ko dc hỏi mình nhé
Ta có:A=(5/10-5/12):2+(5/12-2/14):2+...+(5/998-5/1000):2
suy ra A=5/10-5/1000=99/200