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Ta có :\(x=2014\Rightarrow2015=x+1\)
\(\Rightarrow f\left(2014\right)=x^{17}-\left(x+1\right)x^{2016}+\left(x+1\right)x^{2015}-.....+\left(x+1\right)x-1\)
\(=x^{17}-x^{17}-x^{2016}+x^{2016}+x^{2015}-....+x^2+x-1\)
\(=x-1=2014-1=2013\)
\(|2015x-2014|=|2015x+2014|\)
\(\Leftrightarrow\orbr{\begin{cases}-2015x+2014=|2015x+2014|\left(l\right)\\2015x-2014=|2015x+2014|\left(n\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2015x+2014=-2015x+2014\\2015x+2014=2015x-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}4030x=0\\0x=-4028\left(l\right)\end{cases}\Leftrightarrow}4030x=0\Leftrightarrow x=0}\)
a: \(A=\left|x-8\right|+3>=3\)
Dấu '=' xảy ra khi x=8
b: x=2016 nên x-1=2015
\(P=x^{10}-x^9\left(x-1\right)-x^8\left(x-1\right)-...-x\left(x-1\right)-1\)
\(=x^{10}-x^{10}+x^9-x^9+x^8-...-x^2+x-1\)
=x-1=2015
Nếu \(x=2014\Rightarrow x+1=2015\)
Ta có :
\(P\left(x\right)=x^4-2015x^3+2015x^2-2015x+2015\)
\(\Rightarrow P\left(2014\right)=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(\Rightarrow P\left(2014\right)=x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)
\(\Rightarrow P\left(2014\right)=0+0+0+0+1\)
\(\Rightarrow P\left(2014\right)=1\)
Vậy \(P\left(2014\right)=1\)
\(\Leftrightarrow2015x-2014=\left|x-1\right|\)
ĐK: \(\left|x-1\right|\ge0\Leftrightarrow2015x-2014\ge0\Leftrightarrow2015x\ge2014\Leftrightarrow x\ge\frac{2014}{2015}\)
\(\left|x-1\right|=\hept{\begin{cases}x-1\text{ nếu }x-1\ge0\Rightarrow x\ge1\\-x+1\text{ nếu }x< -1\end{cases}}\)
từ đây bà tự xét tr` hợp
x<-1 và x >=1 nha~~(nhớ phải t/m đk)
ta có :\(\dfrac{y+z-2015x}{x}=\dfrac{z+x-2015y}{y}=\dfrac{z+y-2015z}{z}\)
=>\(\left(\dfrac{y+z-2015}{x}+2016\right)=\left(\dfrac{z+x-2015y}{y}+2016\right)=\left(\dfrac{x+y-2015z}{z}+2016\right)\)
(=)\(\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)
*Nếu x+y+z\(\ne\)0
=>\(\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)
=>\(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)=1.1.1=1
*Nếu x+y+z=0
=>x=y=z
=>\(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)=2.2.2=8
Ta có : \(2015=2014+1=x+1\)
- Thay x + 1 = 2015 vào biểu thức f(2014) ta được :
\(f_{\left(2014\right)}=2014^{17}-\left(2014+1\right).2014^{16}+...+\left(2014+1\right).2014-1\)
=> \(f_{\left(2014\right)}=2014^{17}-2014^{17}-2014^{16}+...+2014^2+2014-1\)
=> \(f_{\left(2014\right)}=2014-1=2013\)
Xin lỗi nha.\(x^{10}-2015x^9-2015x^8-2017x^7-...-2015x-1\)