x^2018=x^2
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EW
0
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
7 tháng 12 2018
\(\dfrac{1}{x^2-x+1}+\dfrac{2}{x^2-x+2}+\dfrac{3}{x^2-x+3}+...+\dfrac{2018}{x^2-x+2018}=2018\)
\(\Leftrightarrow\left(\dfrac{1}{x^2-x+1}-1\right)+\left(\dfrac{2}{x^2-x+2}-1\right)+\left(\dfrac{3}{x^2-x+3}-1\right)+...+\left(\dfrac{2018}{x^2-x+2018}-1\right)=0\)
\(\Leftrightarrow\dfrac{1-x^2+x-1}{x^2-x+1}+\dfrac{2-x^2+x-2}{x^2-x+2}+\dfrac{3-x^2+x-3}{x^2-x+3}+...+\dfrac{2018-x^2+x-2018}{x^2-x+2018}=0\)
\(\Leftrightarrow-\left(x^2-x\right)\left(\dfrac{1}{x^2-x+1}+\dfrac{1}{x^2-x+2}+\dfrac{1}{x^2-x+3}+...+\dfrac{1}{x^2-x+2018}\right)=0\)
Ta có: \(\dfrac{1}{x^2-x+1}+\dfrac{1}{x^2-x+2}+...+\dfrac{1}{x^2-x+2018}>0\)
\(\Leftrightarrow x^2-x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 2 2020
Xét 2 khai triển:
\(\left(x+1\right)^{2018}=C_{2018}^0+C_{2018}^1x+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}\)
\(\left(x-1\right)^{2018}=C_{2018}^0-C_{2018}^1x+C_{2018}^2x^2-...+C_{2018}^{2018}x^{2018}\)
Cộng vế với vế:
\(\left(x+1\right)^{2018}+\left(x-1\right)^{2018}=2\left(C_{2018}^0+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}\right)\)
\(\Leftrightarrow C_{2018}^0+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}=\frac{1}{2}\left(x+1\right)^{2018}+\frac{1}{2}\left(x-1\right)^{2018}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}=\frac{\frac{1}{2}\left(x+1\right)^{2018}+\frac{1}{2}\left(x-1\right)^{2018}-2^{2017}}{x-1}=\lim\limits_{x\rightarrow1}\frac{1009\left(x+1\right)^{2017}+1009\left(x-1\right)^{2017}}{1}=1009.2^{2017}\)
NM
1
30 tháng 1 2018
Do x=2017 nên x+1=2018
Với x+1=2018 thì y trở thành
y= x5-(x+1).x4+(x+1).x3-(x+1).x2+(x+1).x-1
= x5- x5-x4+x4+x3-x3-x2+x-1=x-1
Với x=2017, giá trị biểu thức f(x) là
f(2017)=2017-1=2016
Vậy ...
\(x^{2018}=x^2\\ \Rightarrow x^{2018}-x^2=0\\ \Rightarrow x^2\left(x^{2016}-1\right)=0\\\Rightarrow\left[{}\begin{matrix}x^2=0\\x^{2016}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{2016}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
x2018 = x2
x2018 : x2 = 1
x2016 = 1
Vậy x ϵ {1; -1; 0}