\(x\) biết: \(\left(\frac{11}{12}+\frac{11}{12\cdot23}+\frac{11}...">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

5 tháng 4 2016

1.(11/12+11/23+11/23+11/24+...+11/89+11/100)+x=5/3

1.(11/12+11/100)+x=5/3

77/75+x=5/3

x=5/3-77/75

x=16/25
 

5 tháng 4 2016

de ot

=> ( 11/12+ 1/12- 1/23+ 1/23- 1/34+...+ 1/89-1/100 ) +x=5/3

=> (11/12+ 1/12-1/100 ) + x=5/3

=> (11/12+ 11/150) + x=5/3

=>99/100 +x = 5/3

=> x = 5/3 - 99/100

=> x= 203 / 300

K dung NHA

23 tháng 2 2018

=> 11/12 + 1/12 - 1/23 + 1/23 - 1/34 + ..... + 1/89 - 1/100 + x = 5/3

=> 11/12 + 1/12 - 1/100 + x = 5/3

=> 99/100 + x = 5/3

=> x = 5/3 - 99/100 = 203/300

Tk mk nha

5 tháng 7 2018

Trước hết tính tổng :

\(\frac{11}{12}+\frac{11}{12\times13}+...+\frac{11}{89\times100}=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{89}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

Do đó \(\frac{99}{100}+x=\frac{5}{3}\)

Vậy \(x-\frac{5}{3}-\frac{99}{100}=\frac{500-297}{300}=\frac{203}{300}\)

Vậy...

17 tháng 2 2019

bang 112222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222233333333333333333333333356152784327152718452314983254623145652186521865216416524

dung ko??

a) Ta có: \(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)

\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)

\(=\frac{-6}{72}-\frac{189}{72}+\frac{24}{72}\)

\(=-\frac{19}{8}\)

b) Ta có: \(-1,75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)

\(=\frac{-7}{4}+\frac{1}{9}+\frac{37}{18}\)

\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)

\(=\frac{5}{12}\)

c) Ta có: \(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)

\(=\frac{12}{30}+\frac{-40}{30}+\frac{-15}{30}\)

\(=-\frac{43}{30}\)

d) Ta có: \(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)

\(=\frac{3}{12}-\frac{6}{15}+\frac{3}{10}\)

\(=\frac{15}{60}-\frac{24}{60}+\frac{18}{60}\)

\(=\frac{3}{20}\)

e) Ta có: \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)

\(=\frac{93}{11}+\frac{29}{8}-\frac{38}{11}\)

\(=5+\frac{29}{8}=\frac{40}{8}+\frac{29}{8}=\frac{69}{8}\)

f) Ta có: \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(=\frac{4}{9}\cdot\left(-7\right)+\frac{59}{9}\cdot\left(-7\right)\)

\(=\left(-7\right)\cdot\left(\frac{4}{9}+\frac{59}{9}\right)=\left(-7\right)\cdot7=-49\)

g) Ta có: \(\frac{-1}{4}\cdot13\frac{9}{11}-0,25\cdot6\frac{2}{11}\)

\(=\frac{-1}{4}\cdot\frac{152}{11}+\frac{-1}{4}\cdot\frac{68}{11}\)

\(=\frac{-1}{4}\cdot\left(\frac{152}{11}+\frac{68}{11}\right)=-\frac{1}{4}\cdot20=-5\)

h) Ta có: \(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{52}{5}+\frac{43}{23}+\frac{1}{2}-\frac{5}{27}\)

\(=\frac{64584}{6210}+\frac{11610}{6210}+\frac{3105}{6210}-\frac{1150}{6210}\)

\(=\frac{78149}{6210}\)

i) Ta có: \(\frac{3}{8}\cdot27\frac{1}{5}-51\frac{1}{5}\cdot\frac{3}{8}+19\)

\(=\frac{3}{8}\cdot\frac{136}{5}-\frac{3}{8}\cdot\frac{206}{5}+\frac{3}{8}\cdot\frac{152}{3}\)

\(=\frac{3}{8}\cdot\left(\frac{136}{5}-\frac{206}{5}+\frac{152}{3}\right)=\frac{3}{8}\cdot\frac{110}{3}\)

\(=\frac{55}{4}\)

2 tháng 8 2017

 mik ko chép lại đề, mik làm luôn: 

a)  x - \(\frac{31}{36}=\frac{-13}{38}\)

x = \(\frac{-13}{18}+\frac{31}{36}\)

\(x=\frac{5}{36}\)

b)\(2-x-\frac{3}{7}=\frac{9}{-21}\)

\(\frac{11}{7}-x=\frac{3}{7}\)

x = \(\frac{11}{7}-\frac{3}{7}\)

x = 8/7

c) x + 3/11 = 23/44

x = 23/44 - 3/11

x = 1/4

d) \(\frac{1}{12}-x=\frac{-11}{9}\)

x = \(\frac{1}{12}+\frac{11}{9}\)

x = 47/36

e) \(x-\frac{2}{3}=\frac{-17}{3}\)

x= -17/3 + 2/3

x = -5 

f) \(x-\frac{1}{2}=\frac{11}{4}.\frac{3}{11}\)

x - 1/2 = 3/4

x = 3/4 + 1/2 

x = 5/4

g) \(2x+\frac{3}{8}=\frac{-21}{32}.\frac{4}{7}\)

2x + 3/8 = -3 / 8

2x = -3/8 - 3/8 

2x = -9/8

x = -9/8.1/2 

x = -9/16

h) x - \(\frac{x}{3}=\frac{3}{57}.\frac{19}{12}\)

x  - \(\frac{x}{3}=\frac{1}{12}\)

x = \(\frac{1}{12}+\frac{x}{3}\)

x = \(\frac{1+4x}{12}\)

=> 12x = 1+4x

12x - 4x = 1

8x = 1

x = 1/8 

2 tháng 8 2017

Trả lời nhanh gọn lẹ nhé, mình k cho :)