Question

[Công chúa đáng yêu]

cho x,y,z thỏa mãn x+y+z=6 và x^2+y^2+z^2=12. Tính A=3x+4y-2z

Answer 1

[hồ thế bảo]

Ta có:

x2+y2+z2=12;x+y+z=6⇒3(x2+y2+z2)−(x+y+z)2=0⇔3(x2+y2+z2)−(x2+y2+z2+2xy+2xz+2yz)=0⇔2x2+2y2+2z2−2xy−2xz−2yz=0⇔(x−y)2+(y−z)2+(z−x)2=0(1)x2+y2+z2=12;x+y+z=6⇒3(x2+y2+z2)−(x+y+z)2=0⇔3(x2+y2+z2)−(x2+y2+z2+2xy+2xz+2yz)=0⇔2x2+2y2+2z2−2xy−2xz−2yz=0⇔(x−y)2+(y−z)2+(z−x)2=0(1)

Mà (x−y)2,(y−z)2,(z−x)2≥0,∀x,y,z(x−y)2,(y−z)2,(z−x)2≥0,∀x,y,z

⇒(x−y)2+(y−z)2+(z−x)2≥0,∀x,y,z⇒(x−y)2+(y−z)2+(z−x)2≥0,∀x,y,z

→(1)→(1) đúng chỉ khi dấu bằng xảy ra

(x−y)2=(y−z)2=(z−x)2=0⇔x−y=y−z=z−x=0⇔x=y=z(x−y)2=(y−z)2=(z−x)2=0⇔x−y=y−z=z−x=0⇔x=y=z

Mà x+y+z=6x+y+z=6⇒x=y=z=2⇒x=y=z=2\, suy ra A=10