已知a^2+a-3=0,求a^4+2a^3-a-1的值
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解：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^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
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a^2+a-3=0
a(a+1)=3
a^4+2a^3-a-1
=a^4+a^3+a^3-a-1
=a^3(a+1)+a(a^2-1)-1
=a^2*a(a+1)+a(a+1)(a-1)-1
=a^2*3+3*(a-1)-1
=3a^2+3a-3-1
=3a(a+1)-4
=3*3-4
=5
a(a+1)=3
a^4+2a^3-a-1
=a^4+a^3+a^3-a-1
=a^3(a+1)+a(a^2-1)-1
=a^2*a(a+1)+a(a+1)(a-1)-1
=a^2*3+3*(a-1)-1
=3a^2+3a-3-1
=3a(a+1)-4
=3*3-4
=5
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