a,(a+b)-(-a+b-c)+(c-a-b)
=a+b+a-b+c+c-a-b
=(a+a-a)-(b+b-b)+2c
=a-b+2c
b, a.(b-c)-a.(b+d)
=a.b-a.c-a.b+a.d
=(a.b-a.b)+(-a.c+-a.d)
= 0 + -a.(c+d)
a, (a+b) - (-a+b-c)+(c-a-b) = a-b+2c
*Xét : (a+b) - (-a+b-c) + (c-a-b)
= a+b+a-b+c+c-a-b
= (a+a-a) - (b+b-b) + (c+c)
= a-b+2c
Vì a-b+2c = a-b+2c
\(\Rightarrow\)(a+b) - (-a+b-c) + (c-a-b) = a-b+2c
Vậy (a+b) - (-a+b-c) + (c-a-b) = a-b+2c
b, a(b-c)-a(b+d) = -a(c+d)
*Xét : a(b-c)-a(b+d)
= ab-ac-ab+ad
= (ab-ab) + [-ac+(-ad)]
= 0 + (-a).(c+d)
= -a(c+d)
Vì -a(c+d) = -a(c+d)
\(\Rightarrow\)a(b-c)-a(b+d) = -a(c+d)
Vậy a(b-c)-a(b+d) = -a(c+d)