giúp mình bài này hj

C

chonhoi110

$mn(m^4 - n^4)$

$=mn[(m^4-1)-(n^4-1)]$

$=mn(m-1)(m+1)(m^2+1)-mn(n-1)(n+1)(n^2+1)$

Ta có:$mn(m-1)(m+1)(m^2+1) \vdots 6$(tích 3 số nguyên)

$mn(n-1)(n+1)(n^2+1) \vdots 6$

\Rightarrow$mn(m-1)(m+1)(m^2+1)-mn(n-1)(n+1)(n^2+1)\vdots 6 (1)$

$mn(m^4 - n^4)$

$= mn(m^2 + 1)(m^2 -1) - mn(n^2 + 1)(n^2 -1)$

$= nm(m - 1)(m +1)(m^2 - 4 + 5) - mn(n - 1)(n +1)(n^2 - 4 + 5)$

$= nm(m - 1)(m +1)(m - 2)(m +2) + 5nm(m - 1)(m +1) - mn(n - 1)(n +1)(n - 2)(n + 2) - 5mn(n - 1)(n +1)$

Chứng minh y như trên, ta có:

$nm(m - 1)(m +1)(m - 2)(m +2) + 5nm(m - 1)(m +1) - mn(n - 1)(n +1)(n - 2)(n + 2) - 5mn(n - 1)(n +1)\vdots 5(2) $

Từ$(1),(2)$\Rightarrow $mn(m^4 - n^4)\vdots 30$
 
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