一直x²-3x-2=0,那么代数式（(x-1)³-x²+1）/(x-1)已知x²-3x-2=0,那么代数式（(x-1)³-x²+1）/(x-1)

一直x²-3x-2=0,那么代数式（(x-1)³-x²+1）/(x-1)已知x²-3x-2=0,那么代数式（(x-1)³-x²+1）/(x-1)
一直x²-3x-2=0,那么代数式（(x-1)³-x²+1）/(x-1)
已知x²-3x-2=0,那么代数式（(x-1)³-x²+1）/(x-1)

一直x²-3x-2=0,那么代数式（(x-1)³-x²+1）/(x-1)已知x²-3x-2=0,那么代数式（(x-1)³-x²+1）/(x-1)
解
x²-3x=2
原式=[(x-1)³-(x²-1)]/(x-1)
=[(x-1)³-(x-1)(x+1)]/(x-1)
=(x-1)²-(x+1)
=x²-2x+1-x-1
=x²-3x
=2