题目内容
化简
(1)4x-2y-6y2-y-3x+5y2;
(2)(x3-2y3-3x2y)-(x3-3y3-4x2y);
(3)2a2-[5a-(
a+2)+2a2].
(1)4x-2y-6y2-y-3x+5y2;
(2)(x3-2y3-3x2y)-(x3-3y3-4x2y);
(3)2a2-[5a-(
1 | 2 |
分析:(1)合并同类项即可求解;
(2)首先去括号,然后合并同类项即可求解;
(3)首先去括号时首先去小括号,然后去中括号,最后合并同类项即可求解.
(2)首先去括号,然后合并同类项即可求解;
(3)首先去括号时首先去小括号,然后去中括号,最后合并同类项即可求解.
解答:解:(1)原式=4x-3x-2y-y-6y2+5y2
=x-3y-y2;
(2)原式=x3-2y3-3x2y-x3+3y3+4x2y
=y3+x2y;
(3)原式=2a2-(5a-
a-2+2a2)
=2a2-5a+
a+2-2a2
=-
a+2
=x-3y-y2;
(2)原式=x3-2y3-3x2y-x3+3y3+4x2y
=y3+x2y;
(3)原式=2a2-(5a-
1 |
2 |
=2a2-5a+
1 |
2 |
=-
9 |
2 |
点评:本题考查了整式的加减,解决此类题目的关键是熟记去括号法则,及熟练运用合并同类项的法则,其是各地中考的常考点.注意去括号法则为:--得+,-+得-,++得+,+-得-.
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