Question

với a,b thuộc Z.Chứng minh rằng:(a+b-c)-(a-b+c)+(b+c-a)-(b-a-c)=2b

Answer 1

Ta có: (a+b-c)-(a-b+c)+(b+c-a)-(b-a-c)

        = a+b-c-a+b-c+b+c-a-b+a+c

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

        =  0+2b+0

        = 2b

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

Answer 2

\(\left(a+b-c\right)-\left(a-b+c\right)+\left(b+c-a\right)-\left(b-a-c\right)\)

\(=a+b-c-a+b-c+b+c-a-b+a+c\)

\(=2b\)