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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;

Đỗ Tuệ Lâm · Answer

<=> $x^2+y^2+\left(xy-3x\right)-\left(3y-3\right)=0$

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

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

$\Leftrightarrow x^2+y^2.3.\left(x-1\right)\left(x+1\right)\left(y-1\right)=0$

=> x = 0 ; 1 ; -1 . y =0 ; 1

P/s : ngủ được ròi:3

Câu trả lời:

<=> $x^2+y^2+\left(xy-3x\right)-\left(3y-3\right)=0$

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

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

$\Leftrightarrow x^2+y^2.3.\left(x-1\right)\left(x+1\right)\left(y-1\right)=0$

=> x = 0 ; 1 ; -1 . y =0 ; 1

P/s : ngủ được ròi:3