Question

已知a²+4a+1=0,

(1)求a³+2a²-7a+2012的值
(2)求(a-1/a)²
(3)a∧4-ma²+1/2a³+ma²+2a=3

Answer 1

a³+2a²-7a+2012
=a(a^2+4a)-2a^2-7a+2012
=-2a^2-8a+2012
=-2(a^2+4a)+2012
=2+2012
=2014 题中用到：a^2+4a=-1

2)
a^2+4a+1=0
显然a≠0
方程两边同除以a
a+1/a=-4
a^2+1/a^2+2=4
a^2+1/a^2=2

(a-1/a)^2
=a^2+1/a^2-2=2=0
(3),a^2=-4a-1
a^3=a^2*a=-4a^2-a=-4(-4a-1)-a=16a+4-a=15a+4
a^4=(a^2)^2=16a^2+8a+1=16(-4a-1)+8a+1=-56a-15

(a^4+ma^2+1)/2a^3+ma^2+2a=3
a^4+ma^2+1=6a^3+3ma^2+6a
2ma^2=-6a^3-6a+a^4+1
-8am-2m=-90a-24-6a-56a-15+1=-152a-38
2m(4a+1)=38(4a+1)
m=19