Câu hỏi của Hồ Thị Minh Châu

Question

Cho x,y,z thỏa mãn x2+y2+z2= xy+yz+xz va x2014+y2014+z2014=3

Tinh GTBT L=x25+y4+z2015

Nguyễn Việt Lâm · Accepted Answer

$x^2+y^2+z^2=xy+xz+yz\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0$

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

$\Rightarrow\left\{{}\begin{matrix}x-y=0\x-z=0\y-z=0\end{matrix}\right.$ $\Rightarrow x=y=z$

$x^{2014}+y^{2014}+z^{2014}=3\Rightarrow3x^{2014}=3\Rightarrow x^{2014}=1$

$\Rightarrow x=y=z=\pm1$

- Nếu $x=y=z=1\Rightarrow L=1+1+1=3$

- Nếu $x=y=z=-1\Rightarrow L=-1+1-1=-1$Câu trả lời: $x^2+y^2+z^2=xy+xz+yz\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0$

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

$\Rightarrow\left\{{}\begin{matrix}x-y=0\x-z=0\y-z=0\end{matrix}\right.$ $\Rightarrow x=y=z$

$x^{2014}+y^{2014}+z^{2014}=3\Rightarrow3x^{2014}=3\Rightarrow x^{2014}=1$

$\Rightarrow x=y=z=\pm1$

- Nếu $x=y=z=1\Rightarrow L=1+1+1=3$

- Nếu $x=y=z=-1\Rightarrow L=-1+1-1=-1$

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