题目内容
一个多项式加上-2x3-3x2y+5y2,得x3-2x2y+3y2.
(1)求这个多项式;
(2)当x=-
,y=1时,求这个多项式的值.
(1)求这个多项式;
(2)当x=-
1 |
2 |
(1)(x3-2x2y+3y2)-(-2x3-3x2y+5y2)
=x3-2x2y+3y2+2x3+3x2y-5y2
=3x3+x2y-2y2,
答:这个多项式为3x3+x2y-2y2.
(2)当x=-
,y=1时,
3x3+x2y-2y2.
=3×(-
)3+(-
)2×1-2×12
=-
+
-2
=-2
.
=x3-2x2y+3y2+2x3+3x2y-5y2
=3x3+x2y-2y2,
答:这个多项式为3x3+x2y-2y2.
(2)当x=-
1 |
2 |
3x3+x2y-2y2.
=3×(-
1 |
2 |
1 |
2 |
=-
3 |
8 |
1 |
4 |
=-2
5 |
8 |
练习册系列答案
相关题目