Giải phương trình \(\frac{x-3}{2014}+\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)
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HT
0
PD
1
6 tháng 3 2018
\(PT\Leftrightarrow\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-1}{1008}-2\right)+\left(\frac{x}{2017}-1\right)\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}=\frac{x-2017}{1008}+\frac{x-2017}{2017}\)
\(\Leftrightarrow\frac{x-2017}{2014}+\frac{x-2017}{2015}-\frac{x-2017}{1008}-\frac{x-2017}{2017}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}\right)=0\)
\(\Rightarrow x=2017\)
NQ
3
30 tháng 6 2020
PT đã cho tương đương với:
\(\left(\frac{x}{2017}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+3}{2014}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}\right)=\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2014}\right)\)
\(\Leftrightarrow x+2017=0\Leftrightarrow x=-2017\)
26 tháng 2 2020
Lời giải:
Tập xác định của phương trình
Sử dụng tính chất tỉ lệ thức, có thể biến đổi phương trình như sau
Lời giải thu được
MN
26 tháng 2 2020
\(\frac{x+2012}{2}+\frac{x+2010}{3}+\frac{x+2011}{5}=\frac{x}{1008}+\frac{x-2}{1009}+\frac{x+1}{2015}\)
\(\Leftrightarrow\frac{x+2012}{2}+\frac{x+2010}{3}+\frac{x+2011}{5}-\frac{x}{1008}-\frac{x-2}{1009}-\frac{x+1}{2015}=0\)
\(\Leftrightarrow\frac{x+2012}{2}+2+\frac{x+2010}{3}+2+\frac{x+2011}{5}+1-\frac{x}{1008}-2-\frac{x-2}{1009}-2-\frac{x+1}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2016}{2}+\frac{x+2016}{3}+\frac{x+2016}{5}-\frac{x+2016}{1008}-\frac{x+2016}{1009}-\frac{x+2016}{2015}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}-\frac{1}{1008}-\frac{1}{1009}-\frac{1}{2015}\right)=0\)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}-\frac{1}{1008}-\frac{1}{1009}-\frac{1}{2015}\ne0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
Vậy tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)
BN
3
CC
11 tháng 2 2020
\(\frac{x}{2012}+\frac{x-1}{2013}+\frac{x-2}{2014}-\frac{x-3}{2015}=\frac{x-4}{1008}\)
\(\Leftrightarrow\left(\frac{x}{2012}+1\right)+\left(\frac{x-1}{2013}+1\right)+\left(\frac{x-2}{2014}+1\right)-\left(\frac{x-3}{2015}+1\right)=\frac{x-4}{1008}+2\)
\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}=\frac{x-4+1008.2}{1008}\)
\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}=\frac{x+2012}{1008}\)
\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}-\frac{x+2012}{1008}=0\)
\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}\ne0\)
\(\Rightarrow x+2012=0\)\(\Leftrightarrow x=-2012\)
Vậy \(x=-2012\)
11 tháng 1 2017
Theo bài ra , ta có :
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)=\left(\frac{x+3}{2013}+1\right)+\left(\frac{x+4}{2012}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+2014}{2014}\right)+\left(\frac{x+1+2015}{2015}\right)=\left(\frac{x+3+2013}{2013}\right)+\left(\frac{x+4+2012}{2012}\right)\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Vì \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)>0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
Vậy \(x=-2016\)
Tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)
Chúc bạn học tốt =))
11 tháng 1 2017
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\frac{x+2}{2014}+1+\frac{x+1}{2015}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\frac{x+2+2014}{2014}+\frac{x+1+2015}{2015}=\frac{x+3+2013}{2013}+\frac{x+4+2012}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\left(x+2016\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
MÀ \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
26 tháng 2 2016
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\) <=> \(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\) <=> \(\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\) <=> \(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\) <=> \(x+2016\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\) <=> x + 2016 = 0 <=> x = -2016
