Các Bác giải giúp em: A = 1/3+1/6 +1/15+...+1/45
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F
22 tháng 1 2020
Bài giải
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b, \(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
M
22 tháng 1 2020
Ta có :
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b,
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
PT
1
7 tháng 8 2015
S=1/2.3+2/3.5+3/5.8+5/8.13+8/13.21+13/21.34
= 1/2-1/3+1/3-1/5+1/5-1/8+1/8-1/13+1/13-1/21+1/21-1/34
= 1/2-1/34=8/17
BD
1
C
1 tháng 10 2023
`4 1/2 xx x - 2 1/2 xx x = 1 1/2`
`=> 9/2 xx x - 5/2 xx x = 3/2`
`=> (9/2-5/2)xx x=3/2`
`=>4/2xx x=3/2`
`=>x=3/2:4/2`
`=>x=3/2:2`
`=>x=3/2xx1/2`
`=>x=3/4`
Vậy `x=3/4`
__
`x xx1/2+x xx1/3=5/6`
`=>x xx(1/2+1/3)=5/6`
`=>x xx(3/6+2/6)=5/6`
`=>x xx5/6=5/6`
`=>x=5/6:5/6`
`=>x=5/6xx6/5`
`=>x=1`
HT
3
19 tháng 1 2018
ta có
(1/3+1/6+1/36) +(1/10+1/15+1/45)+(1/21+1/28)
=(\(\frac{12+6+1}{36}\)+\(\frac{9+6+2}{90}\)+\(\frac{4+3}{84}\)
19/36+17/90+1/12
=(19/36+1/12)+17/90
=7/12+17/90
=105/180+34/180
=139/180
19 tháng 1 2018
1/3 +1/6+1/10+1/15+1/21+1/28+1/36+1/45
=1/1x3+1/3x2+1/2x5+1/3x5+1/3x7+1/7x4+1/4x9+1/9x5
=1/1-1/3+1/3-1/2....+1/9-1/5
=1/1
L
3
HT
10 tháng 9 2016
\(a=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(a=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+\frac{1}{3.5}+...+\frac{1}{5.9}\)
\(a=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
=> \(a=2.\frac{2}{5}\)
=> \(a=\frac{4}{5}\)
10 tháng 9 2016
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right)\cdot\frac{1}{2}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{15}+....+\dfrac{1}{45}\)
Đề bài như này thì ko tính đc nhé e phải như này mới tính đc viết nhầm đề rồi hay sao ý
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+....+\dfrac{1}{45}\)
\(A=2\times(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+....+\dfrac{1}{90})\\\)
\(A=2\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{9\times10}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)
\(A=2\times\dfrac{2}{5}\)
\(A=\dfrac{4}{5}\)