tính nhanh 1.4+4.7+7.10+...........................+2015.2018
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MP
1
H
17 tháng 5 2018
Giải:
\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{2015.2018}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{2015}-\dfrac{1}{2018}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{43}+\dfrac{1}{2015}-\dfrac{1}{2018}\)
\(\Leftrightarrow A=\dfrac{42}{43}+\dfrac{1}{2015}-\dfrac{1}{2018}\)
\(\Leftrightarrow A=0,977240464-\dfrac{1}{2018}\)
\(\Leftrightarrow A=0,9767449238\approx0,98\)
Vậy ...
NP
1
K
27 tháng 6 2015
\(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{31.34}\)
\(E=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{31.34}\right)\)
\(E=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{31}-\frac{1}{34}\right)\)
\(E=\frac{2}{3}\left(1-\frac{1}{34}\right)\)
\(E=\frac{2}{3}.\frac{33}{34}\)
\(E=\frac{11}{17}\)
NM
3
MT
5 tháng 8 2015
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
6 tháng 3 2018
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
11 tháng 2 2018
Ta có :
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=\)\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=\)\(1-\frac{1}{43}\)
\(=\)\(\frac{42}{43}\)
18 tháng 9 2016
\(A=\frac{9}{1.4}+\frac{9}{4.7}+\frac{9}{7.10}+...+\frac{9}{97.100}\)
\(A=9\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)
\(A=9.\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...-\frac{1}{100}\right)\)
\(A=\frac{9}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=3\left(\frac{99}{100}\right)=\frac{297}{100}\)
8 tháng 8 2018
Làm từng phần nha bạn
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)
Đặt \(A+x=\frac{299}{301}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)
\(A=1-\frac{1}{301}\)
\(A=\frac{300}{301}\)
=> \(\frac{300}{301}+x=\frac{299}{301}\)
\(x=\frac{299-300}{301}\)
\(x=-\frac{1}{301}\)
8 tháng 8 2018
\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)
\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=\frac{303}{304}\)
\(A=\frac{505}{304}\)
BH
6
10 tháng 7 2019
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\)
\(=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)
MA
3 tháng 2 2022
1.
`16 + (27 - 7.6 ) - (94 -7 - 27.99)`
`= 16+ 27 - 7.6 - 94 + 7 + 27.99`
`= 16 + 27(99 +1) - 7(6-1) - 94`
`= -78 + 27.100 - 7.5`
`= 2587`
2.
`A = 2/1.4 + 2/4.7 + 2/7.10 +...+ 2/97.100`
`A= 2(1/1.4 + 1/4.7 + 1/7.10 +...+1/97.100)`
`3A = 2 (3/1.4 + 3/4.7 + 3/7.10+...+ 3/97.100)`
`3/2 A = 1 - 1/4 + 1/4 - 1/7 +...+ 1/97 - 1/100`
`3/2A = 1 - 1/100`
`3/2 A= 99/100`
`A= 99/100 : 3/2`
`A=33/50`
Vậy `A= 33/50`
TD
3 tháng 2 2022
1.16+(27-7.6)-(94-7-27.99)=16+27-7.6-94+7+27.99
=(27+27.99)+(27+7-94)+16
=27.100-60+16
=2700-44=2656
2.A=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)
=\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
