Câu hỏi của chuche

Question

Giải PT : $x^2+y^2+xy-3x-3y+3=0$

Nguyễn Việt Lâm · Accepted Answer

$\Leftrightarrow\left(x^2+\dfrac{y^2}{4}+\dfrac{9}{4}+xy-3x-\dfrac{3y}{2}\right)+\dfrac{3}{4}\left(y^2-2y+1\right)=0$

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

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

$\Rightarrow x=y=1$

Câu trả lời:

$\Leftrightarrow\left(x^2+\dfrac{y^2}{4}+\dfrac{9}{4}+xy-3x-\dfrac{3y}{2}\right)+\dfrac{3}{4}\left(y^2-2y+1\right)=0$

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

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

$\Rightarrow x=y=1$

Việt Anh · Answer

y = ((căn bậc hai(3)*i-1)*x-căn bậc hai(3)*i+3)/2;y = -((căn bậc hai(3)*i+1)*x-căn bậc hai(3)*i-3)/2;

Câu trả lời:

y = ((căn bậc hai(3)*i-1)*x-căn bậc hai(3)*i+3)/2;y = -((căn bậc hai(3)*i+1)*x-căn bậc hai(3)*i-3)/2;