Question

Tìm x,y biết \(x^2+y^2-4x+6y+13=0\)

Answer 1

\(x^2+y^2-4x+6y+13=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=0\)

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

Vì: \(\left(x-2\right)^2+\left(y+3\right)^3\ge0\forall x;y\)

=> ''='' xảy ra khi x = 2; y = -3

Vậy.........

Answer 2

Lời giải:
\(x^2+y^2-4x+6y+13=0\)

\(\Leftrightarrow (x^2-4x+4)+(y^2+6y+9)=0\)

\(\Leftrightarrow (x-2)^2+(y+3)^2=0\)

Vì \((x-2)^2; (y+3)^2\ge 0, \forall x,y\Rightarrow (x-2)^2+(y+3)^2\geq 0\)

Dấu "=" xảy ra khi \((x-2)^2=(y+3)^2=0\Leftrightarrow \left\{\begin{matrix} x=2\\ y=-3\end{matrix}\right.\)