a, Ta có : 222 ≡ 1(mod 13) nên 222^333 ≡ 1 (mod 13) 
Và 333^2 ≡ -1 (mod 13) nên 333^222 ≡ -1 (mod 13) 
Cộng lại ta có: 
222^333 + 333^222 ≡ 0 (mod 13) đpcm 

b, 2222 ≡ 3 (mod 7) ; 3³ ≡ -1 (mod 7) ; chú ý: 5555 = 3*1851 + 2 
=> 2222^5555 ≡ 3^5555 ≡ (3³)^1851.3² ≡ (-1)^1851.9 ≡ -9 ≡ -2 ≡ 5 (mod 7) 
5555 ≡ 4 (mod 7) ; 4³ ≡ 1 (mod 7) ; 2222 = 3*740 + 2 
=> 5555^2222 ≡ 4^2222 ≡ (4³)^740.4² ≡ (1).16 ≡ 2 (mod 7) 
vậy: 2222^5555 + 5555^2222 ≡ 5+2 ≡ 0 (mod 7) => đpcm 

( tick đúng cho mink nha)

Là điều phải chứng minh đó

b, 5555\(\equiv\)4 (mod 7)=>5555²²²²\(\equiv\)4²²²² (mod 7)⁽¹⁾

2222\(\equiv\)3 (mod 7)=>2222=-4 (mod 7)=>2222⁵⁵⁵⁵\(\equiv\)(-4)⁵⁵⁵⁵ (mod 7)⁽²⁾

Từ ⁽¹⁾  và  ⁽²⁾=>5555²²²²+2222⁵⁵⁵⁵\(\equiv\)4²²²²+4⁵⁵⁵⁵ (mod 7)

                     =>5555²²²²+2222⁵⁵⁵⁵\(\equiv\)4²²²² (1-4³³³³) (mod 7)

Ta có:4³ \(\equiv\)1 (mod 7)

=>(4³)¹¹¹¹\(\equiv\)1¹¹¹¹ (mod 7)

=>4^{3333\(\equiv\)}1 (mod 7)

=>-4^{3333\(\equiv\)}-1(mod 7)

=>1-4³³³³\(\equiv\)0 (mod 7)

=> 5555²²²²+2222⁵⁵⁵⁵\(\equiv\)0 (mod 7)

Vậy 5555²²²²+2222⁵⁵⁵⁵\(⋮\)7

C=đền bài

Ta có:2222 +4 hia hết cho 7 suy ra 2222=-4 (mod7)

suy ra :2222\(^{55555}\)=(-4)\(^{5555}\)(mod7) 55555-4 chia hết cho 7 suy ra 5555=4(mod7)

suy ra 55555\(^{2222}\)=4\(^{2222}\)(mod7)

suy ra 2222\(^{55555}\)5555\(^{2222}\)=(-4)\(^{5555}\)+4\(^{2222}\)(mod7)

mà 4\(^{2222}\)=(-4)\(^{2222}\) suy ra (-4)\(^{5555}\)+4\(^{2222}\)= tự lm típ nha bn mẹt quá

1)=2222^5555 +4^5555 +5555^2222 -4^2222-(4^5555 -4^2222)
=(2222+4).M +(5555-4).N -(4^3333.4^2222 -4^2222)
=(2222+4).M +(5555-4).N -4^2222(4^3333-1)
==(2222+4).M +(5555-4).N --4^2222 (64^1111-1)
==(2222+4).M +(5555-4).N -4^2222(63K)
ta thấy 2222+4=2226 chia hết 7
5555-4 =5551 chia hết cho 7
63 chia hết cho 7
-=>(2222^5555) + (5555^2222) chia hết cho 7

2)Ban viet de sai phai la A=7.5^2+12.6^n=7.25^n+12.6^n =7.25^n-7.6^n+19.6^n =7.(25^n-6^n)+19.6^n Vi 25^n-6^n chia het 19 (do 25-6=19 chia het 19) va 19.6^n chia het 19 => A chia het 19

3¹³=3¹⁰.3³=(3²)⁵.27=9⁵.27=....9.27=.....3

Vậy ..................

a ) 5

b ) 1

0 vs 0

thanks trc