Question

已知：a=根号3 +1，求a^3+a^2-4a-7

已知：a=根号3 +1，求a^3+a^2-4a-7
最好是方法，例如因式分解什么的。因为我已用计算器算出答案是-9
谢谢
楼下是对的是根号3 -1

Answer 1

∵a=根号3 -1
∴a+1=√3
∴a²+2a+1=3
∴a²=2-2a=2（1-a)
a^3+a^2-4a-7
=a²(a+1)-4a-7
=2(1-a)（1+a)-4a-7
=2(1-a²)-4a-7
=2(1-2+2a)-4a-7
=-9

Answer 2

(a-1)^3
= a^3-3a^2+3a - 1
= (a^3+ a^2 - 4a - 7) - 4a^2 + 7a +6
= (a^3+ a^2 - 4a - 7) - 4(a-1)^2 - a +10
(a^3+ a^2 - 4a - 7)
=(a-1)^3 + 4(a-1)^2 + a -10
= 3√3 + 4(3) + √3+1 -10
= 4√3 + 3

Answer 3

a=√3+1
a-2=√3-1
a(a-2)=3-1=2
a^2-2a=2

a^3+a^2-4a-7
=a^3-2a^2+3a^2-4a-7
=a^3-2a^2+3a^2-6a+2a-7
=a(a^2-2a)+3(a^2-2a)+2a-7
=2a+3×2+2a-7
=4a-1
=4a-4+3
=4(a-1)+3
=4√3+3

Answer 4

错了，是√3-1

a+1=√3
平方
a^2+2a+1=3
a^2+2a=2

原式=a^3+2a^2-a^2-4a-7
=a(a^2-2a)-a^2-4a-7
=2a-a^2-4a-7
=-(a^2+2a)-7
=-2-7
=-9