Question

已知a^2+a-3=0,求a^4+2a^3-a-1=

已知a^2+a-3=0,求a^4+2a^3-a-1=

Answer 1

a²=-a+3
以下a²用-a+3代替
a³=a×a²
=a(-a+3)
=-a²+3a
=-(-a+3)+3a
=4a-3

a^4=(a²)²
=(-a+3)²
=a²-6a+9
=(-a+3)-6a+9
=-7a+12

原式(-7a+12)+2(4a-3)-a-1
=-7a+12+8a-6-a-1
=5

Answer 2

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
=2(a^2+a)-1
=6-1
=5