a) x2+3y2+2z2−2x+12y+4z+15=0x2+3y2+2z2−2x+12y+4z+15=0
(x−1)2+3(y+2)2+2(z+1)2=0(x−1)2+3(y+2)2+2(z+1)2=0
Mà : (x−1)2+3(y+2)2+2(z+1)2≥0(x−1)2+3(y+2)2+2(z+1)2≥0
Dấu "=" xảy ra ⇔x=1,y=−2,z=−1⇔x=1,y=−2,z=−1
Vậy (x,y,z)=(1,−2,−1)(x,y,z)=(1,−2,−1)
b) 3x2+y+z2+2x−2y+2xy+3=03x2+y+z2+2x−2y+2xy+3=0
(x+y−1)2+2(x+1)2+z2=0(x+y−1)2+2(x+1)2+z2=0
Mà : (x+y−1)2+2(x+1)2+z2≥0(x+y−1)2+2(x+1)2+z2≥0
Nên dấu "=" xảy ra ⇔z=0,x=−1,y=2⇔z=0,x=−1,y=2
Vậy (x,y,z)=(−1,2,0)
Chúc bạn học tốt!!!