J小爱
幼苗
共回答了17个问题采纳率:88.2% 举报
kx²-2xy+3y²+3x-5y+2
=kx^2+(3-2y)x+(3y-2)(y-1)
所以,设=kx²-2xy+3y²+3x-5y+2
=(ax+(3y-2))(bx+(y-1))
=abx^2+[b(3y-2)+a(y-1)]x+(3y-2)(y-1)
=abx^2+[(3b+a)y-(2b+a)]x+(3y-2)(y-1)
所以:
ab=k
3b+a=-2
-(2b+a)=3
解方程组得:
a=-5,b=1
k=-5
1年前
7