Vì I x + 2016 I > 0
I x + 2017 I > 0
2018 > 0
=> 3x > 0
=> x > 0
Ta có:
I x + 2016 I + I x + 2017 I + 2018 = 3x
<=> x + 2016 + x + 2017 + 2018 = 3x
<=> ( x + x ) + ( 2016 + 2017 + 2018 ) = 3x
<=> 2x + 6051 = 3x
=> x = 6051
Vậy x = 6051
Hok tốt
Tìm x biết
|x+2016|+|x+2017|+2018=3x
/x+2016/ + / x+ 2017/+2018 =3x
\(|x+2016|+|x+2017|\)+2018=3x
Tìm x biết:
\(\frac{x+2014}{2015}+\frac{x+2015}{2016}=\frac{x+2016}{2017}+\frac{x+2017}{2018}\)
1. Cho A=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)và B=\(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}+\frac{1}{2018}\)Tính \(\left(\frac{A}{B}\right)^{2018}\)
2. Tìm x biết
a)\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
b)\(|x+2016|+|x+2017|+2018=3x\)
cho \(x,y\ne0\)thỏa mãn\(x^{2015}+x^{2015}=x^{2016}+x^{2016}=x^{2017}+x^{2017}\)
tính \(S=2018.\left(x^{2018}+y^{2018}\right)\)
tìm x biết
\(\frac{x+18}{2018}+\frac{x+17}{2017}+\frac{x+16}{2016}\)=3
Tìm các số tự nhiên x , y , z thỏa mãn phương trình : 2016^x+2017^y=2018^z
\(\frac{x-1}{2018}+\frac{x-2}{2017}-\frac{x-3}{2016}=\frac{x-4}{2015}\)
Tìm x