1.已知A=x²-x,B=x²-2x+1,则A+B=(2x²-3x+1 ),A-B=(x-1 )
2.要使代数式2mxy-xy+1-2x²-4mxy+5x²+7xy不含xy项,则m的值为( 3)
3.代数式2x²+ax-y+6与2bx²-3x+5y-1的差与字母x的取值无关,求代数式3分之1a²-3b²-(4分之1a³-2b²)的值。
2x²+ax-y+6-(2bx²-3x+5y-1)=2x²+ax-y+6-2bx²+3x-5y+1=(2-2b)x²+(a+3)x-6y-1
2-2b=0,a+3=0
a=-3,b=1
(a²-3b²)/3-(a³-2b²)/4=(9-3)/3-(-27-2)/4=37/4
3)3x²-4x-5=7,3x²-4x-12=0
x=
1.A+B=x²-x+x²-2x+1=2x²-3x+1,A-B=x²-x-(x²-2x+1)=x²-x-x²+2x-1=x-1
2.要使代数式2mxy-xy+1-2x²-4mxy+5x²+7xy不含xy项,
2mxy-xy+1-2x²-4mxy+5x²+7xy=-2mxy+6xy+3x²+1
m=3
A+B=x2-x+x2-2x+1=2x2-3x+1=(2x-1)(x-1)
A-B=-x+2x-1=X-1