Question

已知a的平方+a-3=0，则a的四次方+2a的三次方-a-1的值是？

请给出完整过程

Answer 1

a²+a-3=0
a(a+1)=3

a^4+2a³-a-1
=a^4+a³+a³+a²-a²-a-1
=a³(a+1)+a²(a+1)-a(a+1)-1
=(a+1)(a³+a²-a)-1
=a(a+1)(a²+a-1)-1
=3[a(a+1)-1]-1
=5

Answer 2

a^2+a-3=0
a^2+a=3
a^4+2a^3-a-1
=a^2*(a^2+a)+a^3-a-1
=a^3+3a^2-a-1
=a*(a^2+a)+2a^2-a-1
=2a^2+2a-1
=2(a^2+a)-1
=5

Answer 3

a^2+a-3=0
a^2+a=3
a^4+2a^3-a-1
=a^2*(a^2+a)+a^3-a-1
=a^3+3a^2-a-1
=a*(a^2+a)+2a^2-a-1
=2a^2+2a-1
=2(a^2+a)-1
=5