Question

Cho x+y+z = 0 và xy+yz+zx= 0. Tính giá trị biểu thức:

\(B=\left(x-1\right)^{2007}+y^{2008}+\left(z+1\right)^{2009}\)

Answer 1

\(x+y+z=0\)

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

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

Mà \(xy+yz+xz=0\)

\(\Rightarrow x^2+y^2+z^2+2.0=0\)

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

Mà \(x^2\ge0\)

\(y^2\ge0\)

\(z^2\ge0\)

\(\Rightarrow x^2+y^2+z^2\ge0\)

Mà \(x^2+y^2+z^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)

\(\Rightarrow B=\left(0-1\right)^{2007}+0^{2008}+\left(0+1\right)^{2009}\)

\(=\left(-1\right)^{2007}+0+1^{2009}\)

\(=-1+0+1\)

\(=0\)

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