Tìm x \(\in\) Z biết:
2) \(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
23 tháng 5 2022
a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
=>16x-7=13x+2
=>3x=9
hay x=3
b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)
=>x+2017=0
hay x=-2017
e: \(\left(2x-3\right)^2=144\)
=>2x-3=12 hoặc 2x-3=-12
=>2x=15 hoặc 2x=-9
=>x=15/2 hoặc x=-9/2
8 tháng 4 2023
a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)
=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)
=>2x-2018<0
=>x<2019
b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)
=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)
=>\(x+97< 0\)
=>x<-97
NH
5 tháng 3 2023
\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
KL
Tìm x \(\in\) Z biết:
1) \(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
1
15 tháng 8 2017
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ...
15 tháng 8 2017
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ....
DH
15 tháng 8 2017
\(\dfrac{1-18x}{2017}+\dfrac{2-18x}{2016}=\dfrac{3-18x}{2015}+\dfrac{4-18x}{2014}\)
\(\Rightarrow\left(\dfrac{1-18x}{2017}+1\right)+\left(\dfrac{2-18x}{2016}+1\right)=\left(\dfrac{3-18x}{2015}+1\right)+\left(\dfrac{4-18x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-18x}{2017}+\dfrac{2018-18x}{2016}-\dfrac{2018-18x}{2015}-\dfrac{2018-18x}{2014}=0\)
\(\Rightarrow\left(2018-18x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow2018-18x=0\Rightarrow x=\dfrac{1009}{9}\)
Vậy.............
Chúc bạn học tốt!!!
HD
15 tháng 8 2017
\(\dfrac{1-18x}{2017}+\dfrac{2-18x}{2016}=\dfrac{3-18x}{2015}+\dfrac{4-18x}{2014}\)
\(\Rightarrow\left(\dfrac{1-18x}{2017}+1\right)+\left(\dfrac{2-18x}{2016}+1\right)=\left(\dfrac{3-18x}{2015}+1\right)+\left(\dfrac{4-18x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-18x}{2017}+\dfrac{2018-18x}{2016}-\dfrac{2018-18x}{2015}-\dfrac{2018-18x}{2014}=0\)
\(\Rightarrow\left(2018-18x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow2018-18x=0\)
\(\Rightarrow18x=2018-0\)
\(\Rightarrow18x=2018\)
\(\Rightarrow x=2018:18\)
\(\Rightarrow x=\dfrac{1009}{9}\)
NH
23 tháng 3 2017
1,
đặt A= \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\)
2A=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\)
2A-A=(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\))-(\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\))
A=1-\(\dfrac{1}{2017}\)
A=\(\dfrac{2016}{2017}\)
vậy A=\(\dfrac{2016}{2017}\)
BO
2
30 tháng 3 2021
Ta có: \(\dfrac{2x-5}{2015}+\dfrac{2x-6}{2016}=\dfrac{2x+3}{2007}+\dfrac{2x+4}{2006}\)
\(\Leftrightarrow\dfrac{2x-5}{2015}+1+\dfrac{2x-6}{2016}+1=\dfrac{2x+3}{2007}+1+\dfrac{2x+4}{2006}+1\)
\(\Leftrightarrow\dfrac{2x+2010}{2015}+\dfrac{2x+2010}{2016}-\dfrac{2x+2010}{2007}-\dfrac{2x+2010}{2006}=0\)
\(\Leftrightarrow\left(2x+2010\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
mà \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
nên 2x+2010=0
\(\Leftrightarrow2x=-2010\)
hay x=-1005
Vậy: S={-1005}
LK
30 tháng 3 2021
\(\dfrac{2x-5}{2015}+\dfrac{2x-6}{2016}=\dfrac{2x+3}{2007}+\dfrac{2x+4}{2006}\)
= \(\left(\dfrac{2x-5}{2015}-1\right)+\left(\dfrac{2x-6}{2016}-1\right)=\left(\dfrac{2x+3}{2007}+1\right)+\left(\dfrac{2x+4}{2006}+1\right)\)
=\(\dfrac{2x-2010}{2015}+\dfrac{2x-2010}{2016}=\dfrac{2x+2010}{2007}+\dfrac{2x+2010}{2006}\)
=\(\dfrac{2x-2010}{2015}+\dfrac{2x-2010}{2016}-\dfrac{2x-2010}{2007}-\dfrac{2x-2010}{2006}\)=0
=(2x-2010)\(\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)\)=0
vì \(\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)\)≠0
⇒ 2x-2010=0
⇔2x=2010
⇔x=1005
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(2018-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(2017>2016>2015>2014\) nên
\(\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\Rightarrow x=1009\)
Vậy...........
Chúc bạn học tốt!!!
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(20418-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(Ta\) \(có\)\(:\) \(2017>2016>2015>2014\)
\(\Rightarrow\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\)
\(\Rightarrow2x=2018-0\)
\(\Rightarrow2x=2018\)
\(\Rightarrow x=2018:2\)
\(\Rightarrow x=1009\)