Toán Lớp 8: Tìm x; y thỏa mãn phương trình sau:
x2 – 4x + y2 – 6y + 15 = 2
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Toán Lớp 8: Tìm x; y thỏa mãn phương trình sau:
x2 – 4x + y2 – 6y + 15 = 2
Comments ( 2 )
x² – 4x + y² – 6y + 15 = 2
⇔x² – 4x + 4+ y² – 6y + 9 + 2 = 2 ( tách 16 thành 4+9+2)
⇔(x² – 4x + 4)+ (y² – 6y + 9 )+ 2 = 2
⇔ (x-2)²+ (y-3)²= 0
vì (x-2)²≥0 vs mọi x, (y-3)²≥0 vs mọi x
nên (x-2)²+ (y-3)²≥ vs mọi x,y
đẳng thức xảy ra chỉ khi
⇒\(\left[ \begin{array}{l}(x-2)²=0\\(y-3)²=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x-2=0\\y-3=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=2\\y=3\end{array} \right.\)
vậy x {2;3}
x^2-4x+y^2-6y+13=0
x^2-4x+4+y^2-6y+9=0
(x-2)^2+(y-3)^2=0
Vì (x-2)^2>=0
(y-3)^2>=0
nên (x-2)^2+(y-3)^2>=0
Dấu “=” khi x-2=0; y-3=0 hay x=2; y=3