tìm x biết :
2018+2017+2016+...+(x+1)+x=2018
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NT
0
L
1
24 tháng 8 2019
Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)
\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))
\(\Leftrightarrow x=1\)
Vạy x=1
5 tháng 5 2017
Ta có: \(N\left(x\right)=x^{2017}-2018x^{2016}+2018x^{2015}-...-2018x^2+2018x-1\)
\(=x^{2017}-2018\left(x^{2016}-x^{2015}+...+x^2-x\right)-1\)
\(\Rightarrow N\left(2017\right)=2017^{2017}-2018\left(2017^{2016}-2017^{2015}+...+2017^2-2017\right)-1\)
Đặt \(A=2017^{2016}-2017^{2015}+...+2017^2-2017\)
\(\Rightarrow2017A=2017^{2017}-2017^{2016}+...+2017^3-2017^2\)
\(\Rightarrow2018A=2017^{2017}-2017\)
\(\Rightarrow A=\dfrac{2017^{2017}-2017}{2018}\)
\(\Rightarrow N\left(2017\right)=2017^{2017}-2018.\dfrac{2017^{2017}-2017}{2018}-1\)
\(=2017^{2017}-\left(2017^{2017}-2017\right)-1\)
\(=2017^{2017}-2017^{2017}+2017-1\)
\(=2016\)
Vậy N(2017) = 2016
30 tháng 9 2017
Lâp bảng xét dấu
2016 2017
x-2016 _ 0 + +
x-2017 _ _ 0 +
Nếu x<2016 thì |x-2016|=2016-x,|x-2017|=2017-x
Ta có 2016-x+2017-x=2018
4033-2x=2018
2x=2015
x=1007,5
Nếu 2016<=x<=2017thif |x-2016|=x-2016;|x-2017|=2017-x
Ta có x-2016+2017-x=2018
ox+1=2018
0x=2017 (vô lí)
Nếu x>=2017 thi |x-2016|=x-2016;|x-2017|=x-2017
Ta có x-2016+x-2017=2018
2x-4033=2018
2x=6051
x=3025,5
Vậy x=1007,5 hoăc x=3025,5
PT
2
24 tháng 8 2017
vì (x-2016)^2016 >= 0 vs mọi x
(y-2017)^2018>= 0 vs mọi y
/x+y-z/ >= 0 vs mọi x,y,z
mà (x-2016)^2016+(y-2017)^2018+/x-y+z/=\(\hept{\begin{cases}\left(x-2016\right)^{2016}=0\\^{\left(-2017\right)^{2018}}=0\\x+y-z=0\end{cases}}\)0 nên \(\hept{\begin{cases}x-2016=0\\y-2017=0\\x+y-z\end{cases}}\)\(\hept{\begin{cases}x=2016\\y=2017\\x+y-z=0\end{cases}}\)
24 tháng 8 2017
mà x+y=2016+2017=4033
\(\Rightarrow\)4033-z=0
z=4033
vậy x=2016 y=2017 z=4033
5 tháng 7 2018
Với x>0
\(\Rightarrow x+2016+x+2017+2018=3x\)
\(\Rightarrow x=2016+2017+2018\)
\(\Rightarrow x=6051\)(t/m)
Với x<0
\(\Rightarrow2016-x+2017-x+2018=3x\)
\(\Rightarrow6051-2x=3x\)
\(\Rightarrow x=\frac{6051}{5}\)(loại)
2018+.....+(x+1)+x=\(\dfrac{\left(2018+x\right)\left(2018-x+1\right)}{2}=2018\)
\(\Leftrightarrow\left(2018+x\right)\left(2019-x\right)=4036\)
\(\Leftrightarrow2018.2019+x-x^2=2018.2\)
\(\Leftrightarrow2018.2017+x-x^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2018\left(l\right)\\x=-2017\left(nh\right)\end{matrix}\right.\)
vậy x=-2017