Ta có:
$9^{2007} = 9^{2000}.9^{7}$
$= \left ( 9^{10} \right )^{200}.9^{5}.9^{2}$
$= \left ( ...01 \right )^{200}.\left ( ...49 \right ).81$
$= \left ( ...01 \right ).\left ( ...49 \right ).81$
$= \left ( ...49 \right ).81$
$= \left ( ...69 \right )$
$7^{2007} = 7^{2004}.7^{3}$
$= \left ( 7^{4} \right )^{501}.343$
$= \left ( ...01 \right )^{501}.343$
$= \left ( ...01 \right ).343$
$= \left ( ...43 \right )$
Vậy $9^{2007} - 7^{2007} = \left ( ...69 \right ) - \left ( ...43 \right ) = \left ( ...26 \right )$
Tức là hiệu trên có chữ số tận cùng là $26$.
