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![](https://rs.olm.vn/images/avt/0.png?1311)
b)(1-3-5+7)+(9-11-13+15)+...+(2009-2011-2013+2015)
=0+0+...+0
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=2015-2013+2011-2009+...+7-5+3-1
\(\Rightarrow\)A= 2+2+2+...+2
Có : (2015-1):2+1= 1008 số hạng
Có : 1008:2=504 cặp
\(\Rightarrow\)A=2x504=1008
Vậy A = 1008
![](https://rs.olm.vn/images/avt/0.png?1311)
1-3+5-7+...+2009-2011+2013-2015
=(2013-2015)+(2009-2011)+.....+(1-3)
=-2+-2+...+-2
=-2.1008
=-2016
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2015+2013+2011+...+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)
Ta có : 2015 + 2013 + 2011 + ... + 5 + 3 + 1
= [(2015 - 1) : 2 + 1].(2015 + 1) : 2
= 1008.2016 : 2 = 1016064
Lại có : 2015 - 2013 + 2011 - 2009 + ... + 7 - 5 + 3 - 1 (1008 số hạng
= (2015 - 2013) + (2011 - 2009) + ... + (7 - 5) + (3 - 1) (504 cặp)
= 2 + 2 + ... + 2 + 2 (504 số hạng 2)
= 2 x 504 = 1008
Khi đó A = \(\frac{1016064}{1008}=1008\)
b) tTa có : B = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
=> \(\frac{B}{100}\) = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{97.3}+\frac{100}{99.1}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+\frac{1}{5}+\frac{1}{95}+..+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{2}\)
Khi đó : B/100 = 1/2
=> B = 50
Vậy B = 50
![](https://rs.olm.vn/images/avt/0.png?1311)
( x - 140) : 7 = 3^ 3 - 2^3. 3
( x - 140) : 7 = 27 - 24
( x - 140) : 7 = 3
( x - 140) = 3.7
( x - 140) = 21
x = 21 + 140
x = 161
(x-140):7=27-24
(x-140):7=3
x-140=21
x=161
Ta có: 2A = 2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/(2011*2013)
= 1 - 1/3 + 1/3 -1/5 + ... + 1/2011 - 1/2013
= 1 - 1/2013 = 2012/2013
=> A = 1006/2013
\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+........+\frac{1}{2011\cdot2013}\)
\(\Rightarrow2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+........+\frac{2}{2011\cdot2013}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{2011}-\frac{1}{2013}\)
\(\Rightarrow2A=1-\frac{1}{2013}=\frac{2012}{2013}\)
\(\Rightarrow A=\frac{1006}{2013}\)