Question

已知a+b=3,b+c=1,求代数式a^2+b^2+c^2+ab+bc-ac的值

如题

Answer 1

解：a^2+b^2+c^2+ab+bc-ac
=1/2 *（2a^2+2b^2+2c^2+2ab+2bc-2ac）
=1/2 *[（a^2+b^2 +2ab）+(a^2+c^2-2ac) +(b^2+c^2+2bc)]
=1/2 *[(a+b)^2+(a-c)^2+(b+c)^2]
又a+b=3,b+c=1
故：a +b -(b+c)=3 -1=2,即：a -c=2
因此：a^2+b^2+c^2+ab+bc-ac
=1/2 *[(a+b)^2+(a-c)^2+(b+c)^2]
=1/2 *[3^2+ 2^2+ 1^2]
=1/2 *（9 +4 +1）=1/2 * 14 =7

Answer 2

因为a+b=3,b+c=1,两式相减得a-c=2
a^2+b^2+c^2+ab+bc-ac=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(9+1+4)
=7

Answer 3

原式=[(a+b)^2+(b+c)^2+(a-c)^2]/2.Y由题设知：a+b=3,b+c=1,a-c=2.故原式=[9+1+4]/2=7.

Answer 4

7