1.(a+b)^2+4(a+b)(a-b)+4(a-b)^2 因式分解，有过程

1.(a+b)^2+4(a+b)(a-b)+4(a-b)^2
=[(a+b)+2(a-b)]²
=(a+b+2a-2b)²
=(3a-b)²

2.已知a=1/20x+20,b=1/20+19, c=1/20+21
a^2+b^2+c^2-ab-bc-ac
=(1/2)(2a²+2b²+2c²-2ab-2bc-2ac)
=(1/2)[(a-b)²+(b-c)²+(a-c)²]
=(1/2)(1+1+4)
=3
答案：B.3

1.(a+b)^2+4(a+b)(a-b)+4(a-b)^2 因式分解，有过程
=(a+b+2a-2b)²
=(3a-b)²

2.已知a=1/20x+20,b=1/20+19, c=1/20+21,那么代数式a^2+b^2+c^2-ab-bc-ac的值是？
A 4. B.3 C .2 D 1
a^2+b^2+c^2-ab-bc-ac
=[(a-b)²+(b-c)²+(c-a)²]/2
=(1+4+1)/2
=3

1(a+b)^2+4(a+b)(a-b)+4(a-b)^2
=[(a+b)+2(a-b)]^2
=(3a-b)^2

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+(c-b)²+(c-a)²]
代进去所以选B