Câu hỏi của Bướm Đêm Sát Thủ

Question

tìm tất cả các số x,y,z nguyên thỏa mãn:$x^2+y^2+z^2-xy-3y-2z+4=0$

Phạm Nguyễn Tất Đạt · Accepted Answer

$x^2+y^2+z^2-xy-3y-2z+4=0$

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

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

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

$\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}y=0\\dfrac{1}{2}y-1=0\z-1=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}y\\dfrac{1}{2}y=1\z=1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}x=1\y=2\z=1\end{matrix}\right.$Câu trả lời: $x^2+y^2+z^2-xy-3y-2z+4=0$

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

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

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

$\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}y=0\\dfrac{1}{2}y-1=0\z-1=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}y\\dfrac{1}{2}y=1\z=1\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}x=1\y=2\z=1\end{matrix}\right.$

Violympic toán 8