\(\frac{2013x}{xy+2013x+2013}+\frac{y}{yz+y+2013}+\frac{z}{xz+z+1}\)

\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)

\(=\frac{xz+z+1}{xz+z+1}=1\)

=>đpcm

2013x/xy+2013x+2013 + y/yz+y+2013 + z/xz+z+1

= xyz.x/xy+xyz.x+xyz + y/yz+y+xyz + z/xz+z+1

= xz/1+xz+z + 1/z+1+xz + z/xz+z+1

= xz+1+x/1+xz+x = 1 (đpcm)

Thay xyz=2013 vào ta có:

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

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

\(=\frac{xy\cdot xz}{xy\left(xz+z+1\right)}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

\(=\frac{xz+1+z}{xz+z+1}=1\) (Đpcm)

Đặt \(g\left(x\right)=x^{2015}-x^{2014}+x^{2013}-...+x-1\)

Dễ thấy: \(f\left(x\right)=x^{2016}-2013\times g\left(x\right)\Rightarrow f\left(2012\right)=2012^{2016}-2013\times g\left(2012\right)\)(a)

Ta có: \(\left(x+1\right)\times g\left(x\right)=\left(x+1\right)\left(x^{2015}-x^{2014}+x^{2013}-...+x-1\right)\)

\(\Rightarrow\left(x+1\right)\times g\left(x\right)=x^{2016}-1\)

\(\Rightarrow\left(2012+1\right)\times g\left(2012\right)=2012^{2016}-1\)hay: \(2013\times g\left(2012\right)=2012^{2016}-1\)

Thay vào (a) ta có: \(f\left(2012\right)=2012^{2016}-\left(2012^{2016}-1\right)=1\).

f(x) = x²⁰¹³ - 2013x²⁰¹² + 2013x²⁰¹¹ - 2013x²⁰¹⁰ + .... + 2013x - 1 

= x²⁰¹³ - (2012 + 1)x²⁰¹² + (2012 + 1)x²⁰¹¹ - (2012 + 1)x²⁰¹⁰ + .... + (2012 + 1)x - 1 

= x²⁰¹³ - (x + 1)x²⁰¹² + (x + 1)x²⁰¹¹ - (x + 1)x²⁰¹⁰ + .... + (x + 1)x - 1 

= x²⁰¹³ - x . x²⁰¹² - 1 . x²⁰¹² + x . x²⁰¹¹ + 1 . x²⁰¹¹ - x . x²⁰¹⁰ - 1 . x²⁰¹⁰ + ... + x . x + 1 . x - 1

= x²⁰¹³ - x²⁰¹³ - x²⁰¹² + x²⁰¹² + x²⁰¹¹ - x²⁰¹¹ - x²⁰¹⁰ + .... + x² + x - 1

= x - 1 = 2012 - 1 = 2011

x,y,z=0

Đặt \(\frac{x}{2011}=\frac{y}{2012}=\frac{z}{2013}=k\)

\(\Rightarrow\hept{\begin{cases}x=2011k\\y=2012k\\z=2013k\end{cases}}\)

+) Ta có : \(\frac{2012z-2013y}{2011}=\frac{2012.2013k-2013.2012k}{2011}=0\)

\(\frac{2013x-2011z}{2012}=\frac{2013.2011k-2011.2013k}{2012}=0\)

\(\frac{2011y-2012x}{2013}=\frac{2011.2012k-2012.2011k}{2013}=0\)

Do đó : \(\frac{2012z-2013y}{2011}=\frac{2013x-2011z}{2012}=\frac{2011y-2012x}{2013}\left(=0\right)\) ( đpcm )

Đặt \(\frac{x}{2011}=\frac{y}{2012}=\frac{z}{2013}=k\Rightarrow\hept{\begin{cases}x=2011k\\y=2012k\\z=2013k\end{cases}}\)

\(\frac{2012z-2013y}{2011}=\frac{2012\cdot2013k-2013k\cdot2012}{2011}=\frac{0}{2011}=0\)(1)

\(\frac{2013x-2011z}{2012}=\frac{2013\cdot2011k-2011\cdot2013k}{2012}=\frac{0}{2012}=0\)(2)

\(\frac{2011y-2012x}{2013}=\frac{2011\cdot2012k-2012\cdot2011k}{2013}=\frac{0}{2013}=0\)(3)

Từ (1) , (2) và (3) => đpcm

\(\sqrt{x-2013}+x^3=\sqrt{y-2013}+y^3\)

\(\Leftrightarrow\sqrt{x-2013}-\sqrt{y-2013}+x^3-y^3=0\)

\(\Leftrightarrow\dfrac{x-y}{\sqrt{x-2013}+\sqrt{y-2013}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x-2013}+\sqrt{y-2013}}+\left(x^2+xy+y^2\right)\right)=0\)

\(\Leftrightarrow x=y\)

\(\Rightarrow B=\dfrac{2013x+2014y}{2013y+2014x}=1\)

Quy đồng vế trái ta có

\(\frac{4026}{x^4+x^2+1}=\frac{2014}{x.\left(x^4+x^2+1\right)}\)

Lại quy đồng 2 vế ta được

\(\frac{4026.x}{x.\left(x^4+x^2+1\right)}=\frac{2014}{x.\left(x^4+x^2+1\right)}\)

Suy ra: 4026.x =2014

<=>\(x=\frac{2014}{4026}\)

rút gọn là xong.OK?