已知整式6x-1的值是2,y²的值是4,则（5x²y+5xy-7x)-(4x²y+5xy-7x)=

题目

已知整式6x-1的值是2,y²的值是4,则（5x²y+5xy-7x)-(4x²y+5xy-7x)=

答案

6x-1=2
x=1/2
y^2-y=2
y^2-y-2=0
(y-2)(y+1)=0
y=2或y=-1
(5x的平方y+5xy-7x)-(4x的平方y+5xy-7x)
=5x^2y+5xy-7x-4x^2y-5xy+7x
=x^2y
x^2y=(1/2)^2*2 =1/2
或x^2y=(1/2)^2*(-1) =-1/4