A=4x^2-2xy+4y^2,B=3x^2-6xy+3y^2,|x|=3,y^2=16,|x+y|=1.求4A+[(2A-B)-3(A+B)]的值
问题描述:
A=4x^2-2xy+4y^2,B=3x^2-6xy+3y^2,|x|=3,y^2=16,|x+y|=1.求4A+[(2A-B)-3(A+B)]的值
答
216
答
由y^2=16得|y|=4,
又|x|=3,|x+y|=1.
所以xy=-12
4A+[(2A-B)-3(A+B)]=3A - 4B=18xy=-216
答
4A+[(2A-B)-3(A+B)]
=4A+[-A-4B]=3A-4B
=3(4x^2-2xy+4y^2)-4(3x^2-6xy+3y^2)
=18xy
,|x|=3,,|x+y|=1,y^2=16
x=3时,y=-4,18xy=-216
x=-3时,y=4,18xy=-216
综上,18xy=-216