Bài $20$.
$a$, $A = ( a + b ) - ( a - b ) + ( a - c ) - ( a + c )$
$⇔ A = a + b - a + b + a - c -a -c$
$⇔ A = (a -a + a - a) + (b +b) - (c+c)$
$⇔ A = 2b - 2c$
$⇔ A = 2(b-c)$
$b$, $B = ( a + b - c ) + ( a - b +c ) - ( b + c - a ) - ( a - b - c )$
$⇔ B = a+b-c+a-b+c-b-c+a-a+b+c$
$⇔ B = (a + a + a - a) + (b - b -b + b) + (-c + c - c + c)$
$⇔ B = 2a$
Bài $22$
$|x| < 2013$
$⇒$ $|x|$ $∈$ `{0;1;2;3;......;2011;2012}` vì $|x| ≥ 0 ∀ x$
$⇒$ $x$ $∈$ `{0;±1;±2;±3;....;±2011;±2012}`
Vậy $x$ $∈$ `{0;±1;±2;±3;....;±2011;±2012}`