若a^2+3a+1=0,求(a^4-3a^2+1)/(a^3+5a^2+1)的值

来源:学生作业帮助网 编辑:作业帮 时间:2017/09/24 21:55:29
若a^2+3a+1=0,求(a^4-3a^2+1)/(a^3+5a^2+1)的值若a^2+3a+1=0,求(a^4-3a^2+1)/(a^3+5a^2+1)的值若a^2+3a+1=0,求(a^4-3a

若a^2+3a+1=0,求(a^4-3a^2+1)/(a^3+5a^2+1)的值
若a^2+3a+1=0,求(a^4-3a^2+1)/(a^3+5a^2+1)的值

若a^2+3a+1=0,求(a^4-3a^2+1)/(a^3+5a^2+1)的值
a^2+3a+1=0 => a^3+3a^2+a=0;a^4+3a^3+a^2=0;2a^2+6a+2=0
=(a^4-3a^2+1-a^4-3a^3-a^2)/(a^3+5a^2+1)
=(-3a^3-4a^2+1)/(a^3+5a^2+1)
=-3+(11a^2+4)/(a^3+5a^2+1)
=-3+(11a^2+4)/(a^3+5a^2+1-a^3-3a^2-a)
=-3+(11a^2+4)/(2a^2-a+1)
=-3+(11a^2+4)/(2a^2-a+1-a2^2-6a-2)
=-3+(11a^2+4)/(-7a-1)
=-3+(33a+7)/(7a+1)
=-3+4+(5a+3)/(7a+1)
=1+(5a+3)/(7a+1)
又a=(-3(+ -)genhao(5))/2
带入得=1+(-9(+ -) 5genhao(5))/(-19(+-)7genhao(5))

不会

4a/(a-4)
分子分母凑已知的式子就好了。

哥们 你好强