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问题: 一元二次方程

已知a^2-3a-2=0求代数式(a-1)^3-a^2+1 / a-1 的值

解答:

[(a-1)^3-a^2+1 ]/ (a-1)

=[(a-1)(a-1)^2 -(a^2-1)]/(a-1)
=[(a-1)(a-1)^2 -(a+1)(a-1)]/(a-1)
=(a-1)[(a-1)^2 -(a+1)]/(a-1)
=(a-1)^2 -(a+1)
=a^2-2a+1-a-1
=a^2-3a

a^2-3a-2=0 ==>a^2-3a=2
所以,代数式的值2