(2x²-ax+1)-( )=4x²-8x+2 (4y²-7y-3)-( )=2y²-3y+8若x²-6x-2的2倍减去一个多项式得4x²-7y-5,则这个多项式为?多项式-1/3a-1/2与-a-b+3的和是?差是?

问题描述:

(2x²-ax+1)-( )=4x²-8x+2 (4y²-7y-3)-( )=2y²-3y+8
若x²-6x-2的2倍减去一个多项式得4x²-7y-5,则这个多项式为?
多项式-1/3a-1/2与-a-b+3的和是?差是?

解1:
设:(2x²-ax+1)-A=4x²-8x+2
A=2x²-ax+1-(4x²-8x+2)
A=2x²-ax+1-4x²+8x-2
A=-2x²+(8-a)x-1
所以:(2x²-ax+1)-[-2x²+(8-a)x-1]=4x²-8x+2
解2:
设:(4y²-7y-3)-A=2y²-3y+8
A=(4y²-7y-3)-(2y²-3y+8)
A=4y²-7y-3-2y²+3y-8
A=2y²-4y-11
所以:(4y²-7y-3)-(2y²-4y-11)=2y²-3y+8
解3:
设这个多项式为A,
依题意和已知,有:
2(x²-6x-2)-A=4x²-7y-5
A=2(x²-6x-2)-(4x²-7y-5)
A=2x²-12x-4-4x²+7y+5
A=-2x²-12x+7Y+1
答:这个多项式是-2x²-12x+7Y+1.
解4:
[-(1/3)a-1/2]-(-a-b+3)
=-(1/3)a-1/2+a+b-3
=(2/3)a+b-7/2
答:差是(2/3)a+b-7/2.