1:当k=_____________时,1/2(k-1)x^-5+8是x的一次多项式.2:[12-6y+( )]-[-5y^2-( )+( )]=2y^2+3y-4

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1:当k=_____________时,1/2(k-1)x^-5+8是x的一次多项式.2:[12-6y+( )]-[-5y^2-( )+( )]=2y^2+3y-4
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1:当k=_____________时,1/2(k-1)x^-5+8是x的一次多项式.2:[12-6y+( )]-[-5y^2-( )+( )]=2y^2+3y-4
1:当k=_____________时,1/2(k-1)x^-5+8是x的一次多项式.2:[12-6y+( )]-[-5y^2-( )+( )]=2y^2+3y-4

1:当k=_____________时,1/2(k-1)x^-5+8是x的一次多项式.2:[12-6y+( )]-[-5y^2-( )+( )]=2y^2+3y-4
1:当k=_____________时,1/2(k-1)x^2-5x+8是x的一次多项式.这个题的结果是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,
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,

2y6^2+3y-4

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,