13x2 + 9y2 - 30x + 12xy + 25 = 0
<=> (9y2 + 12xy + 4y2) + (9x2 - 30x + 25) = 0
<=> (3y + 2x)2 + (3x - 5)2 = 0
Dễ thấy \(\left(3y+2x\right)^2\ge0;\left(3x-5\right)^2\ge0\forall x,y\)
nên \(\left(3y+2x\right)^2+\left(3x-5\right)^2\ge0\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}3y+2x=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{10}{9}\\x=\dfrac{5}{3}\end{matrix}\right.\)
Đúng 1
Bình luận (0)
