Câu hỏi của Nguyễn Huyền Trang

Question

x2 + 4y<span style="vertical-align:baseline;col...

Toru · Accepted Answer

$x^2+4y^2+z^2-2x-6z+8y+14=0\\Leftrightarrow (x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)=0\\Leftrightarrow (x^2-2\cdot x\cdot1+1^2)+[(2y)^2+2\cdot2y\cdot 2+2^2]+(z^2-2\cdot z\cdot3+3^2)=0\\Leftrightarrow (x-1)^2+(2y+2)^2+(z-3)^2=0$

Ta thấy: $\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\left(2y+2\right)^2\ge0\forall y\\left(z-3\right)^2\ge0\forall z\end{matrix}\right.$

$\Rightarrow\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x;y;z$

Mặt khác: $\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0$

nên ta được:

$\left\{{}\begin{matrix}x-1=0\2y+2=0\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\y=-1\z=3\end{matrix}\right.$

Vậy: ...

Câu trả lời:

$x^2+4y^2+z^2-2x-6z+8y+14=0\\Leftrightarrow (x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)=0\\Leftrightarrow (x^2-2\cdot x\cdot1+1^2)+[(2y)^2+2\cdot2y\cdot 2+2^2]+(z^2-2\cdot z\cdot3+3^2)=0\\Leftrightarrow (x-1)^2+(2y+2)^2+(z-3)^2=0$

Ta thấy: $\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\left(2y+2\right)^2\ge0\forall y\\left(z-3\right)^2\ge0\forall z\end{matrix}\right.$

$\Rightarrow\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x;y;z$

Mặt khác: $\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0$

nên ta được:

$\left\{{}\begin{matrix}x-1=0\2y+2=0\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\y=-1\z=3\end{matrix}\right.$

Vậy: ...

Kiều Vũ Linh · Answer

$x^2+4y^2+z^2-2x-6z+8y+14=0$

$\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)=0$

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

Do $\left(x-1\right)^2\ge0;\left(2y+2\right)^2\ge0;\left(z-3\right)^2\ge0$

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

*) $\left(x-1\right)^2=0$

$x-1=0$

$x=1$

*) $\left(2y+2\right)^2=0$

$2y+2=0$

$2y=-2$

$y=-1$

*) $\left(z-3\right)^2=0$

$z-3=0$

$z=3$

Vậy x = 1; y = -1; z = 3