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28 tháng 5 2017

a)

\(A=2\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{2015\cdot2017}\right)\)

\(=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)

\(=1-\dfrac{1}{2017}\)

\(=\dfrac{2017}{2017}-\dfrac{1}{2017}\)

\(=\dfrac{2016}{2017}\)

\(B=\dfrac{2013\cdot2015\cdot2017}{2018\cdot2013\cdot\left(2014+1\right)}\)

\(=\dfrac{2013\cdot2015\cdot2017}{2018\cdot2013\cdot2015}\)

\(=\dfrac{2017}{2018}\)

b)

Ta có:

\(A=\dfrac{2016}{2017}=1-\dfrac{1}{2017}\)

\(B=\dfrac{2017}{2018}=1-\dfrac{1}{2018}\)

\(\dfrac{1}{2017}>\dfrac{1}{2018}\)

\(\Rightarrow1-\dfrac{1}{2017}< 1-\dfrac{1}{2018}\)

\(\Rightarrow A< B\)

Vậy \(A< B\).

Anh làm nhé!!

Bài làm:

a) Tính A và B

\(A=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\right)\\ =\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2015.2017}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\\ =1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)

\(B=\dfrac{2013.2015.2017}{2018.2013.\left(2014+1\right)}\\ =\dfrac{2013.2015.2017}{2018.2013.2015}=\dfrac{2017}{2018}\)

b) So sánh A và B.

Ta có: \(A=\dfrac{2016}{2017}=1-\dfrac{1}{2017}\\ B=\dfrac{2017}{2018}=1-\dfrac{1}{2018}\\ Mà:\dfrac{1}{2017}>\dfrac{1}{2018}\\ =>1-\dfrac{1}{2017}< 1-\dfrac{1}{2018}\\ =>A< B\)

30 tháng 4 2018

https://hoc24.vn/hoi-dap/question/598367.html

11 tháng 7 2017

\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)

\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)

\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)

\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)

\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)

\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)

\(B=\dfrac{4.9.16.100}{3.8.15.99}\)

\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)

\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)

\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)

21 tháng 3 2017

a, đặt đề bài là A

Ta có : A=( 1-1/2+1/2-1/3+...+1/9-1/10).(x-1)+1/10.x=x-9/10

= (1-1/10).(x-1)+1/10.x

= 9/10 .( x-1 )+1/10.x

=1.x-9/10

nên x= 0 hoặc 1

21 tháng 3 2017

với -1 nữa nha

21 tháng 7 2018

a, Đề sai hả bạn ??

b, \(\dfrac{\left(1,16-x\right).5,25}{\left(10\dfrac{5}{9}-7\dfrac{1}{4}\right).2\dfrac{2}{17}}=75\%\)

\(\dfrac{\left(1,16-x\right).5,25}{\left(\dfrac{95}{9}-\dfrac{29}{4}\right).\dfrac{36}{17}}=\dfrac{75}{100}\)

\(\dfrac{\left(1,16-x\right).5,25}{\left(\dfrac{380}{36}-\dfrac{261}{36}\right).\dfrac{36}{17}}=\dfrac{3}{4}\)

\(\dfrac{\left(1,16-x\right).5,25}{\dfrac{119}{36}.\dfrac{36}{17}}=\dfrac{3}{4}\)

\(\dfrac{\left(1,16-x\right).5,25}{7}=\dfrac{3}{4}\)

=> \(\left[\left(1,16-x\right).5,25\right].4=3.7\)

\(\left[\left(1,16-x\right).5,25\right].4=21\)

( 1,16 - x ) . 5,25 = 21/4

1,16 - x = 21/4 : 5,25

1,16 - x = 1

x = 1,16 - 1

x = 0,16

Vậy x = 0,16

c, \(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{19.21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{19.21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(\dfrac{1}{2}.\dfrac{20}{21}.420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(200-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(0,4.\left(7,5-2,5x\right):0,25=200-212\)

\(0,4.\left(7,5-2,5x\right):0,25=-12\)

0,4 . ( 7,5 - 2,5x ) = -12 . 0,25

0,4 . ( 7,5 - 2,5x ) = -3

7,5 - 2,5x = -3 :0,4

7,5 - 2,5x = -7,5

2,5x = 7,5-(-7,5)

2,5x = 15

x = 6

Vậy x = 6

Vậy x = 51

23 tháng 7 2018

câu a chắc mk nhìn ko rõ vì mk cận mà ko đeo kính, ghi sai đề

6 tháng 7 2017

\(A=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{99.101}\right)\)

\(A=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}....\dfrac{10000}{99.101}\)

\(A=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}....\dfrac{100^2}{99.101}\)

\(A=\dfrac{2.3.4...100}{1.2.3....99}.\dfrac{2.3.4....100}{3.4.5....101}\)

\(A=100.\dfrac{2}{101}=\dfrac{200}{101}\)

Vậy A = \(\dfrac{200}{101}\)

Chúc học tốt!!

24 tháng 4 2017

cho minh xin yeu cau de bai

26 tháng 4 2017

trả hiểu yêu cầu đề bài là j cả

22 tháng 4 2017

(1+1/1.3)...(1+1/2016.2018)

=(1.3+1/1.3)...(2016.2018+1/2016.2018)

=(2.2/1.3)...(2017.2017/2016.2018)

=(2...2017).(2..2017)/(1.2.....2016).(3...2018)

=2017.2/2018

=2017.2/1006.2

=2017/1006

thông cảm nha :))

1 tháng 8 2018

\(a)\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)

\(=\dfrac{5^{2007}\left(5^3+5^5+5^7\right)}{5^{2007}\left(5^4+5^2+1\right)}=\dfrac{5^3+5^5+5^7}{5^4+5^2+1}\)

\(=\dfrac{125+3125+78125}{625+25+1}=\dfrac{81375}{651}=125\)

\(b)-\dfrac{7}{45}+\dfrac{1}{4}+\dfrac{3}{5}+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}+\dfrac{5}{9}\)

\(=\dfrac{-7.52+1.585+3.468+1.195+2.780+1.60-5.260}{2340}\)

\(=\dfrac{-364+585+1404+195+1560+60-1300}{2340}\)

\(=\dfrac{2140}{2340}=\dfrac{107}{117}\)

1 tháng 8 2018

câu a còn cách nào khác ko bn