Tìm x biết:
2x+4/2015 - 2x+4/2016 = 2x+4/2017 - 2x+4/2018
Em cảm ơn trước ạ!😄😄
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6 tháng 4 2020
\(\frac{2x-4}{2014}+\frac{2x-2}{2016}\) và \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)
VT = \(\frac{2x-4}{2014}+\frac{2x-2}{2016}\)
= \(\frac{2x-4}{2014}+1+\frac{2x-2}{2016}+1\)
= \(\frac{2x-2018}{2014}+\frac{2x-2018}{2016}\)
VP = \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)
= \(\frac{2x-1}{2017}+1+\frac{2x-3}{2015}+1\)
= \(\frac{2x-2018}{2017}+\frac{2x-2018}{2015}\)
Mà \(\frac{2x-2018}{2014}>\frac{2x-2018}{2015}\) và \(\frac{2x-2018}{2016}>\frac{2x-2018}{2017}\)
nên \(\frac{2x-4}{2014}+\frac{2x-2}{2016}\) > \(\frac{2x-1}{2017}+\frac{2x-3}{2015}\)
Chúc bn học tốt!!
20 tháng 5 2022
a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)
\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)
=>2x+10=0
hay x=-5
b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)
=>x-2017=0
hay x=2017
\(\dfrac{2x+4}{2015}-\dfrac{2x+4}{2016}=\dfrac{2x+4}{2017}-\dfrac{2x+4}{2018}\)
\(\Rightarrow\left(2x+4\right)\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\left(2x+4\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
Vì \(\dfrac{1}{2015}-\dfrac{1}{2016}\ne\dfrac{1}{2016}-\dfrac{1}{2017}\) nên 2x + 4 = 0
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
Vậy, x = -2
\(\dfrac{2x+4}{2015}-\dfrac{2x+4}{2016}=\dfrac{2x+4}{2017}-\dfrac{2x+4}{2018}\)
\(\Rightarrow\left(2x+4\right)\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\left(2x+4\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
Vì \(\dfrac{1}{2015}-\dfrac{1}{2016}\ne\dfrac{1}{2016}-\dfrac{1}{2017}\) nên \(2x+4=0\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
Vậy, x = -2