初一暑假作业求解

2x^2-4xy+5y^2-12y+24
=(2x^2-4xy+2y^2)+(3y^2-12y+12)+12
=2(x^2-2xy+y^2)+3(y^2-4y+4)+12
=2(x-y)^2+3(y-2)^2+12
因为2(x-y)^2≥0,3(y-2)^2≥0
所以当x=y=2时,
f(x,y)=2x^2-4xy+5y^2-12y+24的最小值是12

f(x,y)=2x^2-4xy+5y^2-12y+24
=(2x-y)^2+(2y-3)^2+15
(2x-y)^2>=0
(2y-3)^2>=0
f(x,y)的最小值=15

2x^2-4xy+5y^2-12y+24
=(2x^2-4xy+2y^2)+(3y^2-12y+12)+12
=2(x-y)^2+3(y-2)^2+12
当x-y=0,y-2=0时,
多项式f(x,y)=2x^2-4xy+5y^2-12y+24有最小值
即2(x-y)^2+3(y-2)^2+12=12