toán casio

T

thienluan14211

Gọi 3 số tự nhiên là (a-1);a;(a+1)
Ta có $(a-1)^3+a^3+(a+1)^3$
$=a^3+3a+a^3+a^3+3a$
$=3a^3+6a$
$=3(a^3+2a)$
$=3a(a^2+2)$
$=3a(a^2-1+3)$
$=3a[(a+1)(a-1)+3]$
$=3(a-1)a(a+1)+9a$
mà $3(a-1)a(a+1) \vdots 3.6=18 => \vdots 9$
$9a \vdots 9$
=>DFCM
 
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