K
Khách

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

13 tháng 7 2017

\(D=12\cdot\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{22}+...+\frac{1}{97}-\frac{1}{202}\right)\)

\(D=12\cdot\left(\frac{1}{6}-\frac{1}{202}\right)\)

\(D=12\cdot\frac{49}{303}\)

\(D=\frac{588}{303}\)

11 tháng 4 2017

kb đc 0

11 tháng 4 2017

2 câu đầu tôi làm đc

30 tháng 8 2018

Help me :((

30 tháng 8 2018

\(B=1\frac{6}{41}\cdot\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}\div\frac{4+\frac{4}{15}+\frac{4}{4}+\frac{4}{2013}}{5+\frac{5}{15}+\frac{5}{4}+\frac{5}{2013}}\right)\cdot\frac{124242423}{237373735}\)

\(B=\frac{47}{41}\cdot\left[\frac{12\left(1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}\right)}{3\left(1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}\right)}\div\frac{4\left(1+\frac{1}{15}+\frac{1}{4}+\frac{1}{2013}\right)}{5\left(1+\frac{1}{15}+\frac{1}{4}+\frac{1}{2013}\right)}\right]\cdot\frac{123}{235}\)

\(B=\frac{47}{41}\cdot\left[\frac{12}{3}\div\frac{4}{5}\right]\cdot\frac{123}{235}\)

\(B=\frac{3}{5}\cdot3\cdot\frac{5}{4}\)

\(B=\frac{9}{4}\)

18 tháng 6 2019

Trả lời

b)(1/3+12/67+13/41)-(79/67-28/41)

=1/3+12/67+13/41-79/67+28/41

=1/3+(12/67-79/67)+(13/41+28/41)

=1/3+(-67/67)+41/41

=1/3+(-1)+1

=1/3+0

=1/3.

18 tháng 6 2019

c)38/45-(8/45-17/51-3/11)

=38/45-8/45+17/51+3/11

=30/45+1/3+3/11

=2/3+1/3+3/11

=3/3+3/11

=1+3/11

=1 3/11.

15 tháng 3 2018

thà chết đi còn hơn làm cái đống này mất gianroi

3 tháng 4 2017

A = 1 - 1/2014 = 2013/2014 

Mà 2013/2014 > 7/12 nên A > 7/12

Làm tắt thông cảm

3 tháng 4 2017

Ko hiểu?

a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)

\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)

\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)

b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)

\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)

\(\simeq40.39\)

11 tháng 7 2018

\(\frac{1}{2013}x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}=2\)

\(\frac{1}{2013}x+1+(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013})=2\)

\(\frac{1}{2013}x+1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\)

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

\(\frac{1}{2013}x+1+1-\frac{1}{2013}=2\)

\(\frac{1}{2013}x-\frac{1}{2013}+2=2\)

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

\(\frac{1}{2013}.\left(x-1\right)=0\)

=> x - 1 = 0

x = 1

11 tháng 7 2018

\(\frac{1}{2013}x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}=2\)

\(\frac{1}{2013}x+\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\)

\(\frac{1}{2013}x+\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\)

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

\(\frac{1}{2013}x+\frac{2012}{2013}=2\)

\(\frac{1}{2013}x=2-\frac{2012}{2013}\)

\(\frac{1}{2013}x=\frac{2014}{2013}\)

\(x=\frac{2014}{2013}:\frac{1}{2013}\)

=> x=2014