(1)(a+2)(______)=a2-4
(2)(______)(5-x)=25-x2
(3)(2a+4b)(______)=16b2-4a2
(4)(xn+yn)(______)=x2n-y2n
(5)(______)(______)=169x2-196y2
(6)(m2-5n)(5n+m2)=(______)
(7)(a+b-c-d)(a-b-c+d)=[(______)+(______)][(______)-(______)]
解:(1)(a+2)(a-2)=a2-4;
(2)(5+x)(5-x)=25-x2;
(3)(2a+4b)(-2a+4b)=16b2-4a2;
(4)(xn+yn)(xn-yn)=x2n-y2n;
(5)(13x-16y)(13x+16y)=169x2-196y2;
(6)(m2-5n)(5n+m2)=m4-25n2;
(7)(a+b-c-d)(a-b-c+d)=[(a-c)+(b-d)][(a-c)-(b-d)].
故答案為:(1)a-2;(2)5+x;(3)-2a+4b;(4)xn-yn;(5)13x-16y;13x+16y;(6)m4-25n2;(7)a-c;b-d;a-c;b-d.
分析:原式各項利用平方差公式計算即可得到結果.
點評:此題考查了平方差公式,熟練掌握公式是解本題的關鍵.