TV
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TT
0
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
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
Tìm x nha mấy bạn :<
\(\frac{x-1}{2015}+\frac{x-3}{2013}=\frac{x-5}{2011}+\frac{x-7}{2009}\)
=> \(\frac{x-1}{2015}-1+\frac{x-3}{2013}-1=\frac{x-5}{2011}-1+\frac{x-7}{2009}-1\)
=> \(\frac{x-2016}{2015}+\frac{x-2016}{2013}=\frac{x-2016}{2011}+\frac{x-2016}{2009}\)
=> \(\frac{x-2016}{2009}+\frac{x-2016}{2011}-\frac{x-2016}{2013}-\frac{x-2016}{2015}=0\)
=> \(\left(x-2016\right).\left(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\right)\)
Vì \(\frac{1}{2009}>\frac{1}{2013};\frac{1}{2011}>\frac{1}{2015}\)
=> \(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\ne0\)
=> \(x-2016=0\)
=> \(x=2016\)