Question

初一数学题,完全平方公式

a=-2004 b=2003 c=-2002
求a2+b2+c2+ab+bc-ac

Answer 1

解：原式=1/2(2a^2+2b^2+2c^2+2ab+2bc-2ac)
=1/2[(a^2+2ab+b^2)+(b^2+2bc+c^2)+(a^2-2ac+c^2)]
=1/2[(a+b)^2+(b+c)^2+(a-c)^2)]
=1/2[(2004+2003)^2+(2003-2002)^2+(2004+2002)^2]
=1/2(2007^2+1^2+2006^2)
=1/2(4028049+1+4024036)
=4026043

Answer 2

a^2+b^2+c^2+ab+bc-ac
=1/2[2a^2+2b^2+2c^2+2ab+2bc-2ac]
=1/2[(a+b)^2+(a-c)^2+(b+c)^2]
=1/2[(-2004+2003)^2+(-2003-(-2002))^2+(2003+ -2002)^2]
=1/2[1+4+1]
=3

Answer 3

(a+b)2+(b+c)2+(a-c)2
----------------------=.......
2

Answer 4

=(b+1)^2+b^2+(b-1)^2-(b+1)b-b(b-1)-(b+1)(b-1)
=b^2+2b+1+b^2+b^2-2b+1-b^2-b-b^2+b-b^2+1
=3