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20 tháng 4 2019

\(M=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2011.2013}\)

\(M=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\right)\)

\(M=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(M=2.\left(1-\frac{1}{2013}\right)\)

\(M=2.\frac{2012}{2013}\)

\(M=\frac{4024}{2013}\)

~Học tốt~

20 tháng 4 2019

M=2.(2/1.3 + 2/3.5 +2/5.7+...+2/2011.2013)

M=2.(1-1/3 +1/3-1/5 +1/5-1/7+... +1/2011-1/2013)

M=2.(1-1/2013)

M=2.2012/2013

M=4024/2013

26 tháng 4 2017

\(\frac{4}{1.3}\)+\(\frac{4}{3.5}\)+\(\frac{4}{5.7}\)+\(\frac{4}{7.9}\)+...+\(\frac{4}{2011.2013}\)

= 1+\(\frac{1}{3}\)-\(\frac{1}{3}\)+\(\frac{1}{5}\)-\(\frac{1}{5}\)+\(\frac{1}{7}\)-\(\frac{1}{7}\)+\(\frac{1}{9}\)+...+\(\frac{1}{2011}\)+\(\frac{1}{2013}\)

=1+       0          +        0        +        0         +...+        0          +         \(\frac{1}{2013}\)

=1+\(\frac{1}{2013}\)

=\(\frac{2014}{2013}\)

k dùm nha

26 tháng 4 2017

\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{2011\cdot2013}\)

\(=2\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2011\cdot2013}\right)\)

\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=2\cdot\left(1-\frac{1}{2013}\right)\)

\(=2\cdot\frac{2012}{2013}\)

\(=\frac{4024}{2013}\)

Câu 2:

\(D=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)

Câu 3: 

\(E=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{205}-\dfrac{1}{207}\right)\)

\(=2\cdot\left(1-\dfrac{1}{207}\right)=2\cdot\dfrac{206}{207}=\dfrac{412}{207}\)

Câu 5: 

\(G=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{16}{17}=\dfrac{4}{17}\)

10 tháng 7 2016

mình sẽ ủng hộ bạn có câu trả lời đúng nhất nhé

25 tháng 4 2018

Ta có : 

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}\)

\(A=\frac{25}{17}\)

Vậy \(A=\frac{25}{17}\)

Chúc bạn học tốt ~ 

25 tháng 4 2018

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}\)

\(A=\frac{25}{17}\)

\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)

\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(B=\frac{21}{4}.33.\frac{4}{21}\)

\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)

\(B=33\)

\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(C=\frac{1}{2}.\frac{98}{99}\)

\(C=\frac{49}{99}\)

8 tháng 3 2020

a)5.(x+3)-2.(x+4)-(x-2)=17 

=> 5x + 15 - 2x - 8 - x + 2 = 17

=> 2x + 9 = 17

=> 2x = 8

=> x = 4

b)  S=1.3+2.4+3.5+...+48.50+49.51

= 1(2 + 1) + 2(3 + 1) + 3(4 + 1) + ... + 48(49 + 1) + 49(50 + 1)

= 1 + 1.2 + 2 + 2.3 + 3 + 3.4 + ... + 48 + 48.49 + 49 + 49.50

= (1 + 2 + 3 + ... + 49) + (1.2 + 2.3 + 3.4 + ... + 49.50)

đặt A = 1 + 2 + 3 + ... + 49 = (1 + 49).49 : 2 = 1225

đặt B = 1.2 + 2.3 + 3.4 + ... + 49.50

3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3

3B = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

3B = 49.50.51

B = 49.50.51 : 3 = 41650

A + B = S = 41650 + 1225 = 42875

31 tháng 3 2017

M = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)4/1.3 + 4/3.5 + 4/5.7 + ... + 4/2015.2017

M = \(2.\frac{2}{1.3}+2.\frac{2}{3.5}+2.\frac{2}{5.7}+...+2.\frac{2}{2015.2017}\) 2 . 2/1.3 + 2 . 2/3.5 + 2 . 2/5.7 + ... + 2 . 2/2015.2017

M = 2 . ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2015.2017 )

M = 2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2015 - 1/2017 )

M = 2 . ( 1 - 1/2017 )

M = 2 . 2016/2017

M = 4032/2017

31 tháng 3 2017

\(M=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(M=2\left(1-\frac{1}{2017}\right)\)

\(M=\frac{4032}{2017}\)

a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

    =1-1/101

    =100/101

b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5

    =(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5

    =(1-1/101).2,5

    =100/101.2,5

    =250/101

c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2

    =(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2

    =(1/2-1/2010).2

    =1004/1005