K
Khách

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

12 tháng 5 2018

(1/12+3 1/6-30,75).x -8 = (3/5+0,415+1/200):0,01

(1/12+19/6-123/4).x-8=(3/5+83/200+1/200):1/100

-55/2.x-8=51/50:1/100

-55/2.x-8=102

-55/2.x=102+8=110

x=110:-55/2=-4

13 tháng 5 2018

Bạn không làm được bài 2 phần A à?

11 tháng 4 2015

a)          ta có công thức \(\frac{a}{n.\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)

ta có \(N=\frac{5^2}{5.10}+\frac{5^2}{10.15}+...+\frac{5^2}{2005.2010}\)

\(N=5\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2005.2010}\right)\)

 \(N=5\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)(sử dụng quy tắc dấu ngoặc)

\(N=5\left[\frac{1}{5}-\left(\frac{1}{10}-\frac{1}{10}\right)-\left(\frac{1}{15}-\frac{1}{15}\right)-...-\left(\frac{1}{2005}-\frac{1}{2005}\right)-\frac{1}{2010}\right]\)

\(N=5\left[\frac{1}{5}-0-0-...-0-\frac{1}{2010}\right]\)

\(N=5\left[\frac{1}{5}-\frac{1}{2010}\right]\)

\(N=5.\frac{401}{2010}\)

\(N=\frac{401}{402}\)

b)         \(M=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)

               ta thấy      \(\frac{1}{11}=\frac{1}{11}\)

                                \(\frac{1}{12}

18 tháng 4 2018

\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)

\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)

\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)

\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)

\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)

\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)

\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)

\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)

\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)

\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)

\(A=\frac{907,425}{28561}-\frac{19}{37}\)

\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)

\(A=\frac{-509084,275}{1056757}=-0,04604282...\)

Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!

12 tháng 6 2020

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)

\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=1+\frac{1}{99}+1+\frac{1}{98}+1+\frac{1}{95}\)

\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{100}{99}+\frac{99}{98}+\frac{96}{95}\)

\(\Leftrightarrow\left(\frac{x-1}{99}-\frac{100}{99}\right)+\left(\frac{x-2}{98}-\frac{99}{98}\right)+\left(\frac{x-5}{95}-\frac{96}{95}\right)=0\)

\(\Leftrightarrow\frac{x-101}{99}+\frac{x-101}{98}+\frac{x-101}{95}=0\)

\(\Leftrightarrow\left(x-101\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=0\)

\(\Leftrightarrow x-101=0\)

\(\Leftrightarrow x=101\)

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)

\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=1+\frac{1}{99}+1+\frac{1}{98}+1+\frac{1}{95}\)

\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{100}{99}+\frac{99}{98}+\frac{96}{95}\)

\(\Leftrightarrow\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}-\frac{100}{99}-\frac{99}{98}-\frac{96}{95}=0\)

\(\Leftrightarrow\left(\frac{x-1}{99}-\frac{100}{99}\right)+\left(\frac{x-2}{98}-\frac{99}{98}\right)+\left(\frac{x-5}{95}-\frac{96}{95}\right)=0\)

\(\Leftrightarrow\frac{x-101}{99}+\frac{x-101}{98}+\frac{x-101}{95}=0\)

\(\Leftrightarrow\left(x-101\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=0\)

Do \(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\ne0\)

Mà \(x-101=0\Leftrightarrow x=101\)

Vậy x = 101 

1 tháng 8 2018

a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)

  =\(\frac{3}{8}+\frac{2}{13}\)

  =\(\frac{55}{104}.\)

b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)

  =\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)

  =\(\frac{2}{7}+\frac{2}{7}\)

  =\(\frac{4}{7}\)

c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)

  =\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)

  =\(-\frac{16}{55}--\frac{1}{50}\)

  =\(-\frac{149}{550}.\)

d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)

  =\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)

  =\(-\frac{417}{1564}\)

e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)

  =\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)

  =\(\frac{380}{391}.\)

g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)

  =\(-\frac{5}{28}+-\frac{132}{85}\)

  = \(-1.731512605.\)

k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................

5 tháng 8 2018

THANK  Ngô Bùi Hoa  làm cho mình bài 2 với 

9 tháng 4 2019

\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)

\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)

\(=-\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)

\(=-\frac{1.2....99}{2.3...100}.\frac{3.4....101}{2.3...100}\)

\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}\)

Học good

9 tháng 4 2019

\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)

\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)

\(=-\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{99.101}{100^2}\)

\(=-\frac{1.2...99}{2.3...100}\cdot\frac{3.4...101}{2.3.100}\)

\(=-\frac{1}{100}\cdot\frac{101}{2}\)

\(=-\frac{101}{200}\)

11 tháng 4 2017

kb đc 0

11 tháng 4 2017

2 câu đầu tôi làm đc

13 tháng 4 2018

\(\frac{8}{9}\)

13 tháng 4 2018

\(M=\frac{8}{9}\)

đúng đấy bạn