计算:(1);(2)(-4ax)2(5a2-3ax2);(3)3x(2x+1)-(2x+1)2;(4)(m-n+2)(m+n-2).

发布时间:2020-08-07 06:17:52

计算:
(1);
(2)(-4ax)2(5a2-3ax2);
(3)3x(2x+1)-(2x+1)2;
(4)(m-n+2)(m+n-2).

网友回答

解:(1)
=-3+1+
=-3+1+2
=0;

(2)(-4ax)2(5a2-3ax2)
=16a2x2?5a2-16a2x2?3ax2
=80a4x2-48a3x4;

(3)3x(2x+1)-(2x+1)2
=(2x+1)[3x-(2x+1)]
=(2x+1)(x-1)
=2x2-x-1;

(4)(m-n+2)(m+n-2)
=[m-(n-2)][m+(n-2)]
=m2-(n-2)2
=m2-n2+4n-4.
解析分析:(1)利用负整数指数幂的性质以及零指数幂的性质直接化简得出
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