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\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
\(\dfrac{x-2}{2018}=\dfrac{x-3}{2017}=\dfrac{x-4}{2016}=\dfrac{x-5}{2015}\)
\(\dfrac{x-2}{2018}+\dfrac{x-3}{2017}=\dfrac{x-4}{2016}+\dfrac{x-5}{2015}\)
\(\left(\dfrac{x-2}{2018}-1\right)+\left(\dfrac{x-3}{2017}-1\right)=\left(\dfrac{x-4}{2016}-1\right)+\left(\dfrac{x-5}{2015}-1\right)\)
\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}=\dfrac{x-2020}{2016}+\dfrac{x-2020}{2015}\)
\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}-\dfrac{x-2020}{2016}-\dfrac{x-2020}{2015}=0\)
\(\left(x-2020\right)\left(\dfrac{1}{2018}+\dfrac{1}{2017}-\dfrac{1}{2016}-\dfrac{1}{2015}\right)=0\)
\(\dfrac{1}{2018};\dfrac{1}{2017};\dfrac{1}{2016};\dfrac{1}{2015}>0\)
Nên \(x-2020=0\)
\(x=0+2020\)
\(x=2020\)
Vậy x bằng 2020
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Rightarrow\dfrac{x-1}{2017}+\dfrac{x-2}{2016}-\dfrac{x-3}{2015}-\dfrac{x-4}{2014}=0\)
\(\Rightarrow\dfrac{x-1}{2017}-1+\dfrac{x-2}{2016}-1-\dfrac{x-3}{2015}+1-\dfrac{x-4}{2014}+1=0\)
\(\Rightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)-\left(\dfrac{x-3}{2015}-1\right)-\left(\dfrac{x-4}{2014}-1\right)=0\)
\(\Rightarrow\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}-\dfrac{x-2018}{2015}-\dfrac{x-2018}{2014}=0\)
\(\Rightarrow x-2018.\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
Để \(x-2018.\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow x-2018=0\)
\(x=2018\)
Ta có :
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Leftrightarrow\)\(\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\) ( trừ 2 vế cho 2 )
\(\Leftrightarrow\)\(\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}=\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}\)
\(\Leftrightarrow\)\(\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}-\dfrac{x-2018}{2015}-\dfrac{x-2018}{2014}=0\)
\(\Leftrightarrow\)\(\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
Nên \(x-2018=0\)
\(\Rightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
\(\dfrac{x+4}{2015}+\dfrac{x+3}{2016}=\dfrac{x+2}{2017}+\dfrac{x+1}{2018}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2015}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+1}{2018}+1\right)\)
\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}=\dfrac{x+2019}{2017}+\dfrac{x+2019}{2018}\)
\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}-\dfrac{x+2019}{2017}-\dfrac{x+2019}{2018}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
Mà \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)
\(\Leftrightarrow x+2019=0\)
\(\Leftrightarrow x=-2019\)
Vậy...
Giải:
\(\dfrac{x+2015}{5}+\dfrac{x+2016}{4}=\dfrac{x+2017}{3}+\dfrac{x+2018}{2}\)
\(\Leftrightarrow2+\dfrac{x+2015}{5}+\dfrac{x+2016}{4}=2+\dfrac{x+2017}{3}+\dfrac{x+2018}{2}\)
\(\Leftrightarrow\dfrac{x+2015}{5}+1+\dfrac{x+2016}{4}+1=\dfrac{x+2017}{3}+1+\dfrac{x+2018}{2}+1\)
\(\Leftrightarrow\dfrac{x+2015+5}{5}+\dfrac{x+2016+4}{4}=\dfrac{x+2017+3}{3}+\dfrac{x+2018+2}{2}\)
\(\Leftrightarrow\dfrac{x+2020}{5}+\dfrac{x+2020}{4}=\dfrac{x+2020}{3}+\dfrac{x+2020}{2}\)
\(\Leftrightarrow\dfrac{x+2020}{5}+\dfrac{x+2020}{4}-\dfrac{x+2020}{3}-\dfrac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy ...
\(\frac{x}{2015}+\frac{x}{2016}=\frac{x}{2016}+\frac{x}{2017}\)
\(\Rightarrow\frac{x}{2015}+\frac{x}{2016}-\frac{x}{2016}-\frac{x}{2017}=0\)
\(\Rightarrow\frac{x}{2015}-\frac{x}{2017}=0\)
\(\Rightarrow x.\left(\frac{1}{2015}-\frac{1}{2017}\right)=0\)
Mà ta thấy \(\frac{1}{2015}-\frac{1}{2017}\ne0\Rightarrow x=0\)
Vậy \(x=0\)
\(\frac{x}{2015}+\frac{x}{2016}=\frac{x}{2016}+\frac{x}{2017}\)
\(\Leftrightarrow\frac{x}{2015}+\frac{x}{2016}-\frac{x}{2016}-\frac{x}{2017}=0\)
\(\Leftrightarrow x\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
\(\Leftrightarrow x=0\).Do \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
Vậy giá trị của x là x=0
\(=>\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)
=> \(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}\right)=\left(x+2018\right).\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\) = 0
Mà \(\dfrac{1}{2014}>0;\dfrac{1}{2015}>0;\dfrac{1}{2016}>0;\dfrac{1}{2017}>0\)
=>\(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\)
=> \(x+2018=0\)
=>x = 0-2018
=> x = 0+(-2018)
=> x = -2018
\(\left|\dfrac{x}{2015}+\dfrac{x}{2016}\right|=\left|\dfrac{x}{2016}+\dfrac{x}{2017}\right|\)
\(\Rightarrow\left|x\right|.\left|\dfrac{1}{2015}+\dfrac{1}{2016}\right|=\left|x\right|.\left|\dfrac{1}{2016}+\dfrac{1}{2017}\right|\)
\(\Rightarrow\left|x\right|.\left(\dfrac{1}{2015}+\dfrac{1}{2016}\right)=\left|x\right|.\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
Từ đó \(\Rightarrow\)
\(\left|x\right|.\left(\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left|x\right|.\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)=0\)
\(\Rightarrow\left|x\right|.\left[\left(\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\right]=0\)
\(\Rightarrow\left|x\right|=0:\left[\left(\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\right]\)
\(\Rightarrow\left|x\right|=0\Rightarrow x=0\)
Vậy \(x=0\)