Question

请帮我计算下下面的数学题,请写出步骤,谢谢.

a2+4a-1=0 求a2+1/a-2=

Answer 1

a2+4a-1=0

a+4-1/a=0

a-1/a=-4

(a-1/a)^2=16

a^2+1/a^2-2=16

a^2+1/a^2=18

Answer 2

因为a2+4a-1=0 所以两边都除以a(显然a≠0)得a-(1/a)=-4,
a^2+(1/a^)2=a^2-2*a(1/a)+(1/a)^2+2*a(1/a)
=(a-(1/a))^2+2=(-4)^2+2=18

Answer 3

a2 4a-1=0a'=1 b=4 c=-1b2-4a'c=16 4=20a=-b /-根号(b2-4a'c)/2a'解得a=-2 根号5/a=-2-根号5把a=-2 根号5/a=-2-根号5代入a2 1/a-2a2 1/a-2=……其余自己去算吧，太麻烦了不知道对不对