Câu hỏi của Bùi Tấn Sỹ

Question

Biết ( Xo; Yo; Zo) là nghiệm nguyên dương của phương trình X^2+Y^2+Z^2=XY+3Y+2Z-4 .Tính Xo+Yo+Zo

Lightning Farron · Accepted Answer

$x^2+y^2+z^2=xy+3y+2z-4$

$\Leftrightarrow4x^2+4y^2+4z^2=4xy+12y+8z-16$

$\Leftrightarrow4x^2+4y^2+4z^2-4xy-12y-8z+16=0$

$\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(3y^2-12y+12\right)+\left(4z^2-8z+4\right)=0$

$\Leftrightarrow\left(2x-y\right)^2+3\left(y-2\right)^2+4\left(z-1\right)^2=0$

Xảy ra khi $\left\{{}\begin{matrix}2x-y=0\y-2=0\z-1=0\end{matrix}\right.$$\left\{{}\begin{matrix}x=z=1\y=2\end{matrix}\right.$

Suy ra $\left\{{}\begin{matrix}x_0=z_0=1\y_0=2\end{matrix}\right.$$\Rightarrow x_0+y_0+z_0=1+1+2=4$Câu trả lời: $x^2+y^2+z^2=xy+3y+2z-4$

$\Leftrightarrow4x^2+4y^2+4z^2=4xy+12y+8z-16$

$\Leftrightarrow4x^2+4y^2+4z^2-4xy-12y-8z+16=0$

$\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(3y^2-12y+12\right)+\left(4z^2-8z+4\right)=0$

$\Leftrightarrow\left(2x-y\right)^2+3\left(y-2\right)^2+4\left(z-1\right)^2=0$

Xảy ra khi $\left\{{}\begin{matrix}2x-y=0\y-2=0\z-1=0\end{matrix}\right.$$\left\{{}\begin{matrix}x=z=1\y=2\end{matrix}\right.$

Suy ra $\left\{{}\begin{matrix}x_0=z_0=1\y_0=2\end{matrix}\right.$$\Rightarrow x_0+y_0+z_0=1+1+2=4$

Violympic toán 8