题目内容
计算:
(1)5x+4x;
(2)-5x2+
x2;
(3)6x-10x2+12x2-5x;
(4)x3-x2+4-2x3-2+3x2.
(1)5x+4x;
(2)-5x2+
| 1 | 5 |
(3)6x-10x2+12x2-5x;
(4)x3-x2+4-2x3-2+3x2.
分析:根据合并同类项的法则:把同类项的系数相加,所得结果作为系数,字母和字母的指数不变,求解即可.
解答:解:(1)原式=9x;
(2)原式=-
x2;
(3)原式=6x-5x-10x2+12x2=x+2x2;
(4)原式=x3-2x3+3x2-x2+4-2=-x3+2x2+2.
(2)原式=-
| 24 |
| 5 |
(3)原式=6x-5x-10x2+12x2=x+2x2;
(4)原式=x3-2x3+3x2-x2+4-2=-x3+2x2+2.
点评:本题考查了合并同类项,解答本题的关键是掌握合并同类项的法则.
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