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

显示全部楼层 · 2009-3-22 23:15:09

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)/(...

千问 · 2009-3-22 23:15:09

4a/(a-4)分子分母凑已知的式子就好了。...

千问 · 2009-3-22 23:15:09

哥们你好强...

千问 · 2009-3-22 23:15:09

不会...