a^2+a-3=0
a^4+2a^3-a-1=?
a^4+2a^3-a-1
=[(a^2+a)^2]-[a^2+a]-1
=[a^4+2a^3+a^2]-[a^2+a]-1
=9-3-1=5
回答完毕！
a^2+a=3
a^4+2a^3-a-1
=a^4+a^3+a^3-a-1
=a^2(a^2+a)+a^3-a-1
=3a^2+a^3-a-1
=a^3+a^2+2a^2-a-1
=a(a^2+a)+2a^2-a-1
=3a+2a^2-a-1
=2a^2+2a-1
=2(a^2+a)-1
=2*3-1
=5
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
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