Tìm x: ( 1.2.3+2.3.4+...+2015.1016.2017)-(5.6+6.7+...+99.100)= (5+10+..+2000)- 1/1.3 - 1/3.5 - ...- 1/97.99 - 4x
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NM
1
28 tháng 9 2015
A = 1.2. + 2.3 + 3.4 + ... + 99.100
3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 2.3.1 + ... + 99.100.101 - 99.100.98
3A = 99.100.101
3A = 999900
A = 333300
23 tháng 3 2021
A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1-1/100 A=99/100 B= (1/5.6+1/6/7+...+1/101.102).3 B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3 B=(1/5-1/102).3 B=97/170
23 tháng 3 2021
1) Tính
a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
17 tháng 3 2017
a) A = 1.3 +2.4 + 3.5 +...+ 97.99 + 98.100
A = 1(2 + 1) + 2(3+1) + 3(4 + 1) +...+ 98(99+1)
= (1.2 + 2.3 + 3.4 +...+ 98.99) + (1 + 2 + 3 +...+ 98)
= [ 1.2.3 + 2.3.(4-1) +...+ 98.99.(100-97)] + [ 1.2 + 2.(3-1) + 3.(4-2) +... 98.(99-97)]
= [ 1.2.3 + 2.3.(4-1) - 1.2.3 + 3.4.(5-2) - 2.3.(4-1) +...+ 98.99.(100-97) - 97.98(99-96)] + [ 1.2 + 2.(3-1) - 1.2 + 3.(4-2) - 2.(3-1) +...+ 98.(99-97) - 97(98-96)]
= 98.99.100:3 + 98.99:2 = 323 400 + 4581 = 328251
17 tháng 3 2017
b) B = 1.2.3 + 2.3.4 + 3.4.5 +...+ 48.49.50
4B = 1.2.3.4 + 2.3.4.(5-1) + 3.4.5.(6-2) +...+ 48.49.50.(51-47)
4B-B = 1.2.3.4 + 2.3.4.(5-1) - 1.2.3.4 + 3.4.5.(6-2) - 2.3.4.(5-1) +...+ 48.49.50.(51-47) - 47.48.49.(50-46)
= 48.49.50.51:4 = 1499400
DT
1
K
27 tháng 6 2015
\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)
\(\frac{99}{20}-2x=\frac{49}{99}\)
\(2x=\frac{99}{20}-\frac{49}{99}\)
\(2x=\frac{8821}{1980}\)
\(x=\frac{8821}{1980}:2\)
\(x=\frac{8821}{3960}\)
29 tháng 7 2015
a)1/5.6+1/6.7+1/7.8+.......+1/99.100
= (1/5-1/6)+(1/6-1/7)+(1/7-1/8)+.....+(1/99-1/100)
= 1/5 - 1/100
= 19/100
b)2/1.3+2/3.5+2/5.7+.........+2/2013.2015
= (1/1-1/3)+(1/3-1/5)+(1/5-1/7)+.....+(1/2013+1/2015)
= 1/1 - 1/2015
= 2014/2015
MT
29 tháng 7 2015
\(a,\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{5}-\frac{1}{100}=\frac{20}{100}-\frac{1}{100}=\frac{19}{100}\)
\(b,\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)
10 tháng 12 2016
Bài 1 : Ta có : a = 1.2 + 2.3 + 3.4 + ....... + 99.100
=> 3a = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...... + 99.100.(101 - 98)
=> 3a = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 99.100.101
=> 3a = 99.100.101
=> a = \(\frac{99.100.101}{3}=333300\)
