\(\Leftrightarrow x^2-4x+4+y^2-6y+9+2=2\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y-3\right)^2=0\)
Dấu '=' xảy ra khi x=2 và y=3
x2−4x+y2−6x+15=2x2−4x+y2−6x+15=2
⇔(x2−4x+4)+(y2−6x+9)−4−9+15−2=0⇔(x2−4x+4)+(y2−6x+9)−4−9+15−2=0
⇔(x−2)2+(y−3)2=0⇔(x−2)2+(y−3)2=0
Lại có :
{(x−2)2≥0(y−3)2≥0{(x−2)2≥0(y−3)2≥0 ∀x,y∀x,y
Dấu "=" xảy ra ⇔x=2;y=3
⇔x2−4x+4+y2−6y+9=0pt⇔x2−4x+4+y2−6y+9=0
⇔(x−2)2+(y−3)2=0⇔(x−2)2+(y−3)2=0
Mà (x−2)2≥0;(y−3)2≥0∀x;y(x−2)2≥0;(y−3)2≥0∀x;y
⇒(x−2)2+(y−3)2≥0⇒(x−2)2+(y−3)2≥0
Dấu == xảy ra ⇔{(x−2)2=0(y−3)2=0⇔{(x−2)2=0(y−3)2=0
⇔{x=2y=3⇔{x=2y=3
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