若多项式3a³-5a²+a+3与多项式3na²-a-1的和没有二次项,则n=多少
问题描述：
若多项式3a³-5a²+a+3与多项式3na²-a-1的和没有二次项,则n=多少
答
3a³-5a²+a+3+3na²-a-1
=3a³+a²(3n-5)-1
∴3n-5=0
∴n=5/3
答
3a³-5a²+a+3+3na²-a-1
=3a³+3na²-5a²+a-a+3-1
=3a³+(3n-5)a²+2
没有二次项
即3n-5=0
n=5/3
答
负三分之五
答
3a³-5a²+a+3+3na²-a-1
-5+3n=0
n=5/3
答
(3a³-5a²+a+3)+(3na²-a-1)
=3a³+a²(-5+3n)-a+2
-5+3n=0
3n=5
n=5/3