Question

[FFPUBGAOVCFLOL]

Cho x+y+z=0 và xy+yz+zx=0

Tính \(T=\left(x-1\right)^{2013}+y^{2013}+\left(z+1\right)^{2013}\)

Answer 1

[Hoàng Nguyễn Văn]

Ta có \(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2=0\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)mà xy+yz+zx=0

\(\Rightarrow x^2+y^2+z^2=0\left(1\right)\)

Lại có: \(x^2,y^2,z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)Kết hợp (1)

\(\Leftrightarrow x^2=y^2=z^2=0\Leftrightarrow x=y=z=0\)

Vậy \(T=\left(0-1\right)^{2013}+0^{2013}+\left(0+1\right)^{2013}=-1+0+1=0\)

Answer 2

[Trí Tiên]

Ta có : \(x+y+z=0\)

\(\Rightarrow\left(x+y+z\right)^2=0\)

\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)

\(\Rightarrow x^2+y^2+z^2=0\) ( Do \(xy+yz+zx=0\) )

\(\Rightarrow x^2+y^2+z^2=xy+yz+zx\)

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

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow x=y=z\)

Khi đó : \(x+y+z=3x=0\)

\(\Rightarrow x=0\Rightarrow x=y=z=0\)

Nên \(T=\left(0-1\right)^{2013}+0^{2013}+\left(0+1\right)^{2013}=0\)

Vậy : \(T=0\).