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\(=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2021}-\dfrac{1}{2024}=\dfrac{1}{4}-\dfrac{1}{2024}=\dfrac{505}{2024}\)
\(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{1}{2021.2024}\)
=\(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{2021}-\)\(\dfrac{1}{2024}\)
=\(\dfrac{1}{4}-\dfrac{1}{2024}\)
=\(\dfrac{505}{2024}\)
S = 1.4 + 4.7 + 7.10 + 10.13 + ... + 61.64
1.4.9 = 1.4.(7 + 2) = 1.4.7 + 1.4.2
4.7.9 = 4.7.(10 - 1) = 4.7.10 - 1.4.7
7.10.9 = 7.10.(13 - 4) = 7.10.13 - 4.7.10
10.13.9 = 10.13.(16 - 7) = 10.13.16 - 7.10.13
.......................................................................
61.64.9 = 61.64.(67 - 58) = 61.64.67 - 58.61.64
Cộng vế với vế ta có:
1.4.9 + 4.7.9 + 7.10.9 +...+ 61.64.9 = 1.4.2 + 61.64.67
9(1.4 + 4.7 + 7.10+ ...+ 61.64) = 261576
1.4 + 4.7 + 7.10 +...+ 61.64 = 261576 : 9
1.4 + 4.7 + 7.10 + ... + 61.64 = 29064
A=6/4.7+6/7.10+6/10.13+...+6/73.76
\(=2.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{76}\right)=2.\frac{9}{38}=\frac{9}{19}\)
\(A=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{25\cdot28}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{6}{28}=\dfrac{2}{28}=\dfrac{1}{14}\)
`3A = 3/(4.7) + 3/(7.10) + .. + 3/(25.28)`
`3A = 1/4 - 1/7 + 1/7 - 1/10 +... + 1/25 - 1/28`
`3A = 3/14`
`A = 1/14.`
a: \(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{1}{3}-\dfrac{1}{203}=\dfrac{200}{609}\)
b: \(B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=\dfrac{1}{4}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)
\(B=1-\dfrac{3}{1\cdot4}-\dfrac{3}{4\cdot7}-...-\dfrac{3}{2020\cdot2023}\\ =1-\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2020\cdot2023}\right)\\ =1-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\\ =1-\left(1-\dfrac{1}{2023}\right)\\ =1-\dfrac{2022}{2023}=\dfrac{1}{2023}\)
`B=1-3/(1.4)-3/(4.7)-3/(7.10)-....-3/(2020.2023)`
`B=1-(3/(1.4)+3/(4.7)+.....+3/(2020.2023))`
`B=1-(1-1/4+1/4-1/7+.....+1/2020-1/2023)`
`B=1-(1-1/2023)`
`B=1-1+1/2023=1/2023`
\(\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+\frac{2}{13\cdot16}\)
\(=2\left(\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}\right)\)
\(=2\left[\frac{1}{3}\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\right)\right]\)
\(=2\left[\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\right]\)
\(=2\left[\frac{1}{3}\left(\frac{1}{4}-\frac{1}{16}\right)\right]\)
\(=2\left[\frac{1}{3}\cdot\frac{3}{16}\right]\)
\(=2\cdot\frac{1}{16}\)
\(=\frac{2}{16}=\frac{1}{8}\)
Ta có :
\(\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}\)
\(=\)\(2\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}\right)\)
\(=\)\(\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)
\(=\)\(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=\)\(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=\)\(\frac{2}{3}.\frac{3}{16}\)
\(=\)\(\frac{1}{8}\)
Chúc bạn học tốt ~
3/1.4+3/4.7+3/7.10+3/10.13
=1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13
=1 - 1/13
=12/13
A=4.7+7.10+10.13+...+205.208
A.9=4.7.9+7.10.9+10.13.9+...+205.208.9
A.9=4.7.(10-1)+7.10.(13-4)+...+205.208.(211-202)
A.9=4.7.10-1.4.7+7.10.13-4.7.10+...+205.208.211-202.205.208
A.9=-1.4.7+205.208.211
A.9=8997012
A=8997012:9=999668
tick cho mình nha