2a^2+3a-1=0,求(2a^5+3a^4+3a^3+9a^2-5a+1)/(3a-1)

来源:百度知道 编辑:UC知道 时间:2024/06/30 11:30:33

先化简分子
2a^5+3a^4+3a^3+9a^2-5a+1
=(2a^5+3a^4-a^3)+(4a^3+6a^2-2a)+(3a^2-3a+1)
=a^3(2a^2+3a-1)+2a(2a^2+3a-1)+(3a^2-3a+1)
=3a^2-(3a-1)
=3a^2-(-2a^2)
=5a^2

分母
3a-1=-2a^2
原式=5a^2/(-2a^2)=-5/2

2a^2+3a-1=0
2a^2+3a=1

2a^2+3a-1=0
3a-1=-2a^2

分子=a^3(2a^2+3a)+3a^3+9a^2-5a+1
=a^3+3a^3+9a^2-5a+1
=4a^3+6a^2+3a^2-5a+1
=2a(2a^2+3a)+3a^2-5a+1
=3a^2-3a+1
=3a^2-(3a-1)
=3a^2+2a^2
=5a^2

分母=-2a^2

所以原式=-5/2