问题
解答题
当a为何值时,多项式3(ax2+2x-2)-(9x2+6x-7)的值恒等于1?
答案
3(ax2+2x-2)-(9x2+6x-7)
=3ax2+6x-6-9x2-6x+7
=(3a-9)x2+1.
依题意得,3a-9=0.
解得a=3.
答:当a=3时,多项式3(ax2+2x-2)-(9x2+6x-7)的值恒等于1.
当a为何值时,多项式3(ax2+2x-2)-(9x2+6x-7)的值恒等于1?
3(ax2+2x-2)-(9x2+6x-7)
=3ax2+6x-6-9x2-6x+7
=(3a-9)x2+1.
依题意得,3a-9=0.
解得a=3.
答:当a=3时,多项式3(ax2+2x-2)-(9x2+6x-7)的值恒等于1.