Trang chủ » Hỏi đáp » Môn Toán Tìm Min ` x^2 + 4y^2 + 9z^2 -4x+12y-24z+30` 09/09/2024 Tìm Min ` x^2 + 4y^2 + 9z^2 -4x+12y-24z+30`
Giải đáp: x^2+4y^2+9z^2-4x+12y-24z+30= (x^2-4x+4)+(4y^2-12y+9)+(9z^2-24z+16)+1= (x^2-2*x*2+2^2)+[(2y)^2-2*2y*3+3^2]+[(3z)^2-2*3z*4+4^2]+1= (x-2)^2+(2y-3)^2+(3z-4)^2+1Vì (x-2)^2\ge0AAx,(2y-3)^2\ge0AAy,(3z-4)^2\ge0AAz=> (x-2)^2+(2y-3)^2+(3z-4)^2\ge0AAx,y,z=> (x-2)^2+(2y-3)^2+(3z-4)^2+1\ge1AAx,y,zDấu “=” xảy ra khi: {(x-2=0),(2y-3=0),(3z-4=0):}<=> {(x=2),(y=3/2),(z=4/3):}Vậy min = 1 <=> {(x=2),(y=3/2),(z=4/3):} Trả lời
= (x^2-4x+4)+(4y^2-12y+9)+(9z^2-24z+16)+1
= (x^2-2*x*2+2^2)+[(2y)^2-2*2y*3+3^2]+[(3z)^2-2*3z*4+4^2]+1
= (x-2)^2+(2y-3)^2+(3z-4)^2+1
Vì (x-2)^2\ge0AAx,(2y-3)^2\ge0AAy,(3z-4)^2\ge0AAz
=> (x-2)^2+(2y-3)^2+(3z-4)^2\ge0AAx,y,z
=> (x-2)^2+(2y-3)^2+(3z-4)^2+1\ge1AAx,y,z
Dấu “=” xảy ra khi: {(x-2=0),(2y-3=0),(3z-4=0):}
<=> {(x=2),(y=3/2),(z=4/3):}
Vậy min = 1 <=> {(x=2),(y=3/2),(z=4/3):}