Question

问一道数学题，求详细解答过程

[(a+b)/2]^2-[(a-b)/2]^2

其中a=-1/8 b=2

先分解因式 然后再求值

Answer 1

[(a+b)/2]²-[(a-b)/2]²
={[(a+b)/2]+[(a-b)/2]}{[(a+b)/2]-[(a-b)/2]}
=(a/2+b/2+a/2-b/2)(a/2+b2-a/2+b/2)
=a*b
=(-1/8)*2
=-1/4
如果不懂请说明

Answer 2

原式=[(a+b)/2+(a-b)/2][(a+b)/2-（a-b)/2]
=ab=-1/8×2=-1/4

Answer 3

[(a+b)/2]²-[(a-b)/2]²
={[(a+b)/2]+[(a-b)/2]}{[(a+b)/2]-[(a-b)/2]}
=(a/2+b/2+a/2-b/2)(a/2+b2-a/2+b/2)
=a*b
=(-1/8)*2
=-1/4

Answer 4

[(a+b)/2]^2-[)(a-b/2]^2=[(a+b)^2-(a-b)^2]/4=[a^2-2ab+b^2-a^2+2ab-b^2]/4=ab=-1/8*2=-1/4

Answer 5

[(a+b)/2]²-[(a-b)/2]²
={[(a+b)/2]+[(a-b)/2]}{[(a+b)/2]-[(a-b)/2]}
=(a/2+b/2+a/2-b/2)(a/2+b2-a/2+b/2)
=a*b
=(-1/8)*2
=-1/4