题目内容
合并同类项
(1)5ab-2ab-3ab=
(2)mn+nm=
(3)-5xn-xn-(-8xn)=
(4)-5a2-a2-(-7a2)+(-3a2)=
(1)5ab-2ab-3ab=
0
0
.(2)mn+nm=
2mn
2mn
.(3)-5xn-xn-(-8xn)=
2xn
2xn
.(4)-5a2-a2-(-7a2)+(-3a2)=
-2a2
-2a2
.分析:(1)利用合并同类项的定义即可求解;
(2)利用合并同类项的定义即可求解;
(3)首先利用符号法则对式子进行化简,然后利用合并同类项的定义即可求解;
(4)首先利用符号法则对式子进行化简,然后利用合并同类项的定义即可求解.
(2)利用合并同类项的定义即可求解;
(3)首先利用符号法则对式子进行化简,然后利用合并同类项的定义即可求解;
(4)首先利用符号法则对式子进行化简,然后利用合并同类项的定义即可求解.
解答:解:(1)5ab-2ab-3ab
=(5-2-3)ab
=0.
(2)mn+nm
=(1+1)mn
=2mn.
(3)-5xn-xn-(-8xn)
=-5xn-xn+8xn
=(-5-1+8)x n
=2xn;
(4)-5a2-a2-(-7a2)+(-3a2)
=-5a2-a2+7a2-3a2
=(-5-1+7-3)a2
=-2a2.
故答案是:0;2mn;2xn;-2a2.
=(5-2-3)ab
=0.
(2)mn+nm
=(1+1)mn
=2mn.
(3)-5xn-xn-(-8xn)
=-5xn-xn+8xn
=(-5-1+8)x n
=2xn;
(4)-5a2-a2-(-7a2)+(-3a2)
=-5a2-a2+7a2-3a2
=(-5-1+7-3)a2
=-2a2.
故答案是:0;2mn;2xn;-2a2.
点评:本题主要考查合并同类项得法则.即系数相加作为系数,字母和字母的指数不变.
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