Ta có  / x-2013/ + /x -2020/ = /x -2013/ + / 2020 -x/ >/   / x -2013 + 2020 -x / = 7

Dấu ' =' xảy ra khi   2013</ x </ 2020

Nếu x là số tự nhiên 

=> x thuộc { 2013; 2014;2015;....2020}

BẰng 13336

13336

QFvƯG

Ta có: a = 2020 => 2021 = x + 1

f(2020) = x²⁰¹⁴ - (x + 1) . x²⁰¹³ + (x + 1) . x²⁰¹² - ... + (x + 1) . x² - (x + 1) . x - 1

= x²⁰¹⁴ - x²⁰¹⁴ + x²⁰¹³ + x²⁰¹³ + x²⁰¹² - ... + x³ + x² - x² + x - 1

= x - 1 = 2020 - 1 = 2019

Vậy f(2020) = 2019

Thay x+y+z=2020 vào \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2020}\) có:

\(\frac{xy+yz+xz}{xyz}=\frac{1}{x+y+z}\)

<=>\(\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)

<=>\(x^2y+xy^2+xyz+xyz+y^2z+yz^2+x^2z+xyz+xz^2=xyz\)

<=>\(xy\left(x+y\right)+z^2\left(x+y\right)+y^2z+x^2z+3xyz-xyz=0\)

<=>\(\left(x+y\right)\left(xy+z^2\right)+z\left(y^2+x^2+2xy\right)=0\)

<=>\(\left(x+y\right)\left(xy+z^2\right)+z\left(x+y\right)^2=0\)

<=>\(\left(x+y\right)\left(xy+z^2+xz+yz\right)=0\)

<=>\(\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]=0\)

<=>\(\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

=> \(\left[{}\begin{matrix}x=-y\\y=-z\\x=-z\end{matrix}\right.\)

Tại x=-y => \(x^{2009}=-y^{2009}\)

<=>\(x^{2009}+y^{2009}\)=0

Có \(P=\left(x^{2009}+y^{2009}\right)\left(y^{2011}+z^{2011}\right)\left(z^{2013}+x^{2013}\right)=0\left(y^{2011}+z^{2011}\right)\left(z^{2013}+x^{2013}\right)=0\)

Tương tự các trường hợp kia cũng => P=0

Vậy P=0

từ giả thiết ta suy ra : \(x,y,z\ne0\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{xz+yz+z^2}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0\Leftrightarrow\left(x+y\right)\left[\left(xz+z^2\right)+\left(yz+xy\right)\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\z+y=0\\x+z=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^{2009}=-y^{2009}\\y^{2011}=-z^{2011}\\z^{2013}=-x^{2013}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^{2009}+y^{2009}=0\\y^{2011}+z^{2011}=0\\z^{2013}+x^{2013}=0\end{matrix}\right.\)

Vậy P = 0

- Xét x^8 - 2014x^7 tại x= 2013

   x^8 - 2014x^7 = 2013^8-2014. 2013^7 = 2013^7. ( 2013-2014) = - 2013^7

- Tính tiếp : -2013^7 + 2014. 2013^6 = 2013^6 ( -2013+2014 ) = 2013^6

Cứ như vậy đến -2014. 2013 = - 2013

Cuối cùng KQ f(2013) = -2013+2020=7

2013^8 - 2014. 2013^7 = 2013^7 ( 2013-2014 ) = - 2013^7

- 2013^7 + 2014. 2013^6 = 2013^6 ( -2013+2014) = 2013^6

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-2013 + 2020 = 7

Vậy f(2013)= 7