Question

1:当k=_____________时,1/2(k-1)x^-5+8是x的一次多项式.2:[12-6y+( )]-[-5

Answer 1

1要没有x^2项 必有k-1=0，k=1

2 :[12-6y+( ay^2)]-[-5y^2-(by )+( c)]=2y^2+3y-4,

则(a+5)y^2-(6+b)y+(12-c)=2y^2+3y-4, 同类项系数相等

a+5=2,a=-3,

-(6+b)=3,b=-9

12-c=-4,c=16

故[12-6y+( -3y^2)]-[-5y^2-(-9y )+( 16)]=2y^2+3y-4,