Question

Cho các số nguyên a;b;c;d: Chứng minh rằng tổng /a-b/+/b-c/+/c-d/+/d -a/ luôn là số chẵn

TKS mn
 

Answer 1

Lời giải:
Đặt $a-b=x; b-c=y; c-d=z; d-a=t$ thì $x+y+z+t=0$

$\Rightarrow x+y=-(z+t)$

$\Rightarrow (x+y)^2=(z+t)^2$

Đặt $A=|a-b|+|b-c|+|c-d|+|d-a|=|x|+|y|+|z|+|t|$

$A^2=(|x|+|y|+|z|+|t|)^2$

$=(|x|+|y|)^2+(|z|+|t|)^2+2(|x|+|y|)(|z|+|t|)$

$=x^2+y^2+z^2+t^2+2|xy|+2|zt|+2(|x|+|y|)(|z|+|t|)$

$=(x+y)^2+(z+t)^2-2xy-2zt+2|xy|+2|zt|+2(|x|+|y|)(|z|+|t|)$

$=2(z+t)^2-2xy-2zt+2|xy|+2|zt|+2(|x|+|y|)(|z|+|t|)$ chẵn

$A^2$ chẵn kéo theo $A$ chẵn (đpcm)