来自胡林献的问题
把(a+b+c+d)(b+c-a-d)(c+a-b-d)(a+b-c-d)+16abcd因式分解为___.
把(a+b+c+d)(b+c-a-d)(c+a-b-d)(a+b-c-d)+16abcd因式分解为___.
1回答
2020-01-16 19:12
把(a+b+c+d)(b+c-a-d)(c+a-b-d)(a+b-c-d)+16abcd因式分解为___.
把(a+b+c+d)(b+c-a-d)(c+a-b-d)(a+b-c-d)+16abcd因式分解为___.
原式=[(a+b)+(c+d)][(c-d)+(b-a)][(c-d)-(b-a)][(a+b)-(c+d)]+16abcd
=[(a+b)2-(c+d)2][(c-d)2-(b-a)2]+16abcd
=(a2+b2-c2-d2+2ab-2cd)(c2+d2-a2-b2-2cd+2ab)+16abcd
=(2ab-2cd)2-(a2+b2-c2-d2)2+16abcd
=4a2b2+4c2d2-8abcd+16abcd-(a2+b2-c2-d2)2
=4a2b2+4c2d2+8abcd-(a2+b2-c2-d2)2
=(2ab+2cd)2-(a2+b2-c2-d2)2
=(2ab+2cd+a2+b2-c2-d2)(2ab+2cd-a2-b2+c2+d2)
=(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d).
故答案为:(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d).