已知a^2-5a+1=0,求代数式a^3-4a^2-4a+3的值

2024-12-26 15:46:20
推荐回答(4个)
回答1:

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

回答2:

a^2-5a+1=0
a²=5a-1

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

回答3:

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

回答4:

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