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CA
7 tháng 6 2018
a,0,36.350+1,2.20.3+9.4.4,5
=13.3.35+12.2.3+9.2.3.3
=3.(13.35+12.2+.9.2.3)
=3.(455+24+54)
=3.533
=1599
b,2015.2016-5/2015.2015+2010
=4062240-5+2010
=4064245
c,2/1.3+2/3.5+2/5.7+...+2/71.73
=1-1/3+1/3-1/5+1/5-1/7+...+1/71-1/73
=1-1/73
=72/73
d,(1+1/2).(1+1/3)+...+(1+1/2018)
=3/2.4/3.5/4+...+2019/2018
=2019/2
e,E=1/4.5+1/5.6+1/6.7+...+1/80.81(làm tương tự với phần d nên mình làm ngắn
=1/4-1/81
=77/324
f,F=3/2.3+3/3.4+...+3/99.100
=3.(1/2.3+1/3.4+...+1/99.100)(làm tương tự với d
=3.(1/2-1/100)
=3.49/100
=147/100
gG=5/1.4+5/4.7+...+5/61.64
3G=5.(3/1.4+3./4.7+...+3/61.64)
=5.(1-1/64)
=5.63/64
=315/64
ok nha bạn,mình giữ đúng lời hứa.
VN
Võ Ngọc Phương
VIP
21 tháng 8 2023
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)
\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=\dfrac{1}{103}\)
\(\Rightarrow x+3=103\)
\(x=103-3\)
\(x=100\)
Vậy x = 100
8 tháng 6 2018
f,F=3. (1/2 .3 + 1/3.4 +...+ 1/99.100)
= 3. (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100
= 3. (1/2 - 1/100)
= 3. 49/100
= 147/100
g, G = 5/3. (3/1.4 + 3/4.7 +...+ 3/61.64)
= 5/3 . (1 - 1/4 + 1/4 - 1/7 +...+ 1/61 - 164
= 5/3 . (1-1/64)
= 5/3 . 63/64
= 105/64
MX
8 tháng 6 2018
f, \(F=\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(\Leftrightarrow F=3\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Leftrightarrow F=3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Leftrightarrow F=3\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Leftrightarrow F=3\left(\frac{49}{100}\right)=\frac{147}{100}\)
g, \(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)
\(\Leftrightarrow G=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{61.64}\right)\)
\(\Leftrightarrow G=5.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{64}\right)\)
\(\Leftrightarrow G=\frac{5}{3}\left(1-\frac{1}{64}\right)\)
\(\Leftrightarrow G=\frac{5}{3}.\frac{63}{64}=\frac{105}{64}\)
8 tháng 6 2018
\(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)
\(\Rightarrow G=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+..+\frac{3}{61.64}\right)\)
\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+..+\frac{1}{61}-\frac{1}{64}\right)\)
\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{64}\right)=\frac{5}{3}.\frac{63}{64}\)
\(\Rightarrow G=\frac{5.63}{3.64}=\frac{5.21.3}{3.64}=\frac{5.21}{64}=\frac{105}{64}\)
BD
3
19 tháng 3 2018
Ta có :
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(A=\frac{2}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(A=\frac{33}{50}\)
Vậy \(A=\frac{33}{50}\)
Chúc bạn học tốt ~
T
19 tháng 3 2018
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
7 tháng 6 2016
a) 1/5.6 + 1/6.7 + 1/7.8 + ... + 1/24.25
= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25
= 1/5 - 1/25
= 4/25
b) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 -1/101
= 1 - 1/101
= 100/101
c) 3/1.4 + 3/4.7 + ... + 3/2002.2005
= 1 - 1/4 + 1/4 - 1/7 + ... + 1/2002 - 1/2005
= 1 - 1/2005
= 2004/2005
d) 5/2.7 + 5/7.12 + ... + 5/1997.2002
= 1/2 - 1/7 + 1/7 - 1/12 + ... + 1/1997 - 1/2002
= 1/2 - 1/2002
= 500/1001
7 tháng 6 2016
a,A = \(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)
A\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
A\(=\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
b, B=\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)
B= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
B=\(1-\frac{1}{101}=\frac{100}{101}\)
c, \(C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{2002\times2005}\)
C= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)
C= \(1-\frac{1}{2005}=\frac{2004}{2005}\)
d, D= \(\frac{5}{2\times7}+\frac{5}{7\times12}+...+\frac{5}{1997\times2002}\)
D= \(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{1997}-\frac{1}{2002}\)
D= \(\frac{1}{2}-\frac{1}{2002}=\frac{1001}{2002}-\frac{1}{2002}=\frac{1000}{2002}=\frac{500}{1001}\)
PQ
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 8 2023
Để olm.vn giúp em nhá
C = \(\dfrac{1}{1.4}\) + \(\dfrac{1}{4.7}\) + \(\dfrac{1}{7.11}\)+...+ \(\dfrac{1}{994.997}\) + \(\dfrac{1}{997.1000}\)
C = \(\dfrac{1}{3}\).( \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.11}\)+...+ \(\dfrac{3}{994.997}\)+ \(\dfrac{3}{997.1000}\))
C = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\)-\(\dfrac{1}{11}\)+...+ \(\dfrac{1}{994}\)- \(\dfrac{1}{997}\)+ \(\dfrac{1}{997}\) - \(\dfrac{1}{1000}\))
C = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{1000}\))
C = \(\dfrac{1}{3}\). \(\dfrac{999}{1000}\)
C = \(\dfrac{333}{1000}\)
12 tháng 8 2015
1/1.4 + 1/4.7 + 1/7.10 + ... + 1/31.34
= 1/3 . ( 3/1.4 + 3/4.7 + 3/7.10 + .... + 3/31.34 )
= 1/3 . ( 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/31 - 1/34 )
= 1/3 . ( 1 - 1/34 )
= 1/3 . 33/34
= 11/34
12 tháng 7 2015
A, 1/4.5 + 1/5.6 + 1/6.7+ 1/7.8
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8
= 1/4 - 1/8
=1/8
B; 3/1.4 + 3/4.7 + 3/7.10 = 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10
= 1- 1/10
= 9/10
C, 1/1.3 + 1/3.5 + 1/5.7 + .. +1/2007.2009
= 1/2 ( 2/1.3 + 2/3.5 + ... + 2/2007.2009)
= 1/2 ( 1/1 - 1/3 + 1/3 - 1/5 + ... +1/2007 - 1/2009)
= 1/2 ( 1- 1/2009)
= 1/2 . 2008/2009
= 1004/2009
D; 8/2.6 + 8/6.10 + 8/10.14 + 8/14.18 + 8/18.22
= 2( 4/2.6 + 4/6.10 + .. +4/18.22)
= 2 ( 1/2 - 1/6 + 1/6 - 1/10 + .. + 1/18 - 1/22)
= 2 ( 1/2 - 1/22)
= 2 .5/11
= 10/11
E; 3/2.4 + 3/4.6 + ... +3/998.1000
= 3/2( 2/2.4 + 2/4.6 +.. +2/998.1000)
= 3/2 .( 1/2 - 1/4 + 1/4 - 1/6 + ... +1/998 - 1/1000 )
=3/2 . 449/1000
= 1497/2000
ẤN đúng cho mình nha làm xong đứt hơi
12 tháng 7 2015
a ) 1/4.5 + 1/5.6 + 1/6.7+ 1/7.8
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8
= 1/4 - 1/8
=1/8
b ) 3/1.4 + 3/4.7 + 3/7.10 = 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10
= 1- 1/10
= 9/10
c ) 1/1.3 + 1/3.5 + 1/5.7 + .. +1/2007.2009
= 1/2 ( 2/1.3 + 2/3.5 + ... + 2/2007.2009)
= 1/2 ( 1/1 - 1/3 + 1/3 - 1/5 + ... +1/2007 - 1/2009)
= 1/2 ( 1- 1/2009)
= 1/2 . 2008/2009
= 1004/2009
d ) 8/2.6 + 8/6.10 + 8/10.14 + 8/14.18 + 8/18.22
= 2( 4/2.6 + 4/6.10 + .. +4/18.22)
= 2 ( 1/2 - 1/6 + 1/6 - 1/10 + .. + 1/18 - 1/22)
= 2 ( 1/2 - 1/22)
= 2 .5/11
= 10/11
e ) 3/2.4 + 3/4.6 + ... +3/998.1000
= 3/2( 2/2.4 + 2/4.6 +.. +2/998.1000)
= 3/2 .( 1/2 - 1/4 + 1/4 - 1/6 + ... +1/998 - 1/1000 )
=3/2 . 449/1000
= 1497/2000
\(E=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{97\cdot100}\)
\(=\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right):3\)
\(=\left(1-\frac{1}{100}\right):3=\frac{33}{100}\)
\(F=\frac{3}{1\cdot5}+\frac{3}{5\cdot9}+...+\frac{3}{74\cdot101}\)
\(=\left(3-\frac{3}{5}+\frac{3}{5}-\frac{3}{9}+...+\frac{3}{74}-\frac{3}{101}\right):4\)
\(=\left(3-\frac{3}{101}\right):4=\frac{75}{101}\)
E = 33/100
F = 75/101