Question

(a+1)(a的平方-2a-3)+(a-1)(a的平方-2)-(2a的平方-1)(a+1),其中a=-1.

请尽快，我在线等

Answer 1

(a+1)(a的平方-2a-3)+(a-1)(a的平方-2)-(2a的平方-1)(a+1)
=(a+1)(a²-2a-3-2a²+1)+(a-1)(a的平方-2)
=(a+1)(-a²-2a-2)+a³-a²-2a+2
=0×(-a²-2a-2)-1-1+2+2
=-2+2+2
=2

Answer 2

(a+1)(a^2-2a-3)+(a-1)(a^2-2)-(2a^2-1)(a+1)
=(a+1)[(a^2-2a-3)-(2a^2-1)]+(a-1)(a^2-2)
=(a+1)(a^2-2a-3-2a^2+1)+(a-1)(a^2-2)
=(a+1)(-a^2-2a-2)+(a-1)(a^2-2)
=-a^3-2a^2-2a-a^2-2a-2+a^3-2a-a^2+2
=-a^3+a^3-2a^2-a^2-a^2-2a-2a-2a+2-2
=-4a^2-6a
=-4*(-1)^2-6*(-1)
=-4+6
=2

Answer 3

2