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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.
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
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
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)
\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)
Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)
\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)
\(2A=\frac{12}{3}-\frac{12}{99}\)
\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)
Tính :
a) \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
c) \(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\frac{2}{75}\)
\(=\frac{1}{75}\)
( \(\frac{1}{1x3}\)+ \(\frac{1}{3x5}\)+....+\(\frac{1}{9x11}\)) x \(y\) = \(\frac{2}{3}\)
( \(\frac{2}{1x3}\)+ \(\frac{2}{3x5}\)+...+\(\frac{2}{9x11}\)) x \(y\) = \(\frac{4}{3}\) (nhân 2 vế lên với 2)
(1 - \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)- ...+ \(\frac{1}{9}\)- \(\frac{1}{11}\)) x \(y\)= \(\frac{4}{3}\)
( 1 - \(\frac{1}{11}\)) x \(y\)=\(\frac{4}{3}\)
\(\frac{10}{11}\) x \(y\) =\(\frac{4}{3}\)
\(y\) = \(\frac{4}{3}\): \(\frac{10}{11}\)
\(y\) = \(\frac{4}{3}\)x \(\frac{11}{10}\)
\(y\) =\(\frac{22}{15}\)
a) A = 3.4 + 4.5 + 5.6 + ...+ 49.50
=> 3A = 3.4.3+4.5.3+ 5.6.3+...+49.60.3
3A = 3.4.(5-2) +4.5.(6-3) + 5.6.(7-4) + ...+ 49.60.(61-48)
3A = 3.4.5 - 2.3.4 + 4.5.6 -3.4.5 + 5.6.7-4.5.6 + 49.60.61 - 48.49.60
3A = -2.3.4 + 49.60.61
\(A=\frac{-2.3.4+49.60.61}{3}=59772\)
b) B = 1.3 + 3.5 + 5.7 + ...+ 51.53
=> 6B = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 51.53.6
6B = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) +...+ 51.53.(55-49)
6B = 1.3.5 + 1.3 + 3.5.6 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 51.53.55 - 49.51.53
6B = 1.3 + 51.53.55
\(B=\frac{1.3+51.53.55}{6}=24778\)
cau c mk ko bk
d) D = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561
D = 30+31+32+33+34+35+36+37+38
=> 3D = 31+32+33+...+38+39
=> 3D - D = 39-30
2D = 39-1
\(D=\frac{3^9-1}{2}=9841\)
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
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= \(\frac{1}{2}-\frac{1}{2002}=\frac{1001}{2002}-\frac{1}{2002}=\frac{1000}{2002}=\frac{500}{1001}\)