a=x⁴-2223x³+2223x²-2223x+2223

=x³(x-2223)+x(x-2223)+2222x²+2003(*)

thay x=2222,ta co:

(*)<=>-2222³-2222+2222³+2223=1

dung thi chon nha

dễ mà                                 

98 x 456 x 123 x 753 x 246 x 2222 x 11 x 11 x 11 x 25 x 36 x 25 x 12 x 13 x 159 x 487 x 258 x 12589 x 1524 x ( 36987 - 36987 ) = 0

:)))

98 x 456 x 123 x 753 x 246 x 2222 x 11 x 11 x 11 x 25 x 36 x 25 x 12 x 13 x 159 x 487 x 258 x 12589 x 1524 x ( 36987 - 36987 ) =0

ht

k mk nha

98 x 456 x 123 x 753 x 246 x 2222 x 11 x 11 x 11 x 25 x 36 x 25 x 12 x 13 x 159 x 487 x 258 x 12589 x 1524 x ( 36987 - 36987 ) 

=>0

ht 

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

mod là j

ai trả lời nhanh và sớm ***

Ta thấy : \(2222^{3333}vs2^{300}:\hept{\begin{cases}2222>2\\3333>300\end{cases}\Rightarrow2222^{3333}>2^{300}}\)

Ta thấy : \(2222^{1111}=1111^{1111}.2^{1111}< 1111^{1111}.1111^{1110}=1111^{2221}\)

Ta thấy : \(54^{10}=\left(3^3\right)^{10}.2^{10}=3^{30}.2^{10}=3^{12}.3^{18}.2^{10}>3^{12}.7^{12}=21^{12}.\)

9/13 x 7/12 + 9/13 x 5/12 - 9/13

= 9/13 x (7/12 + 5/12 - 1)

= 9/13 x 0

= 0

4/13 x 5/12 + 4/13 x 7/12 - 4/3

= 4/13 x (5/12 + 7/12) - 4/3

= 4/13 x 1 - 4/3

= 4/13 - 4/3

= -40/39

a) 9/13 x 7/12 + 9/13 x 5/12 - 9/13

= 9/13 x ( 5/12 + 7/12 - 1 )

= 9/13 x 0

= 0

b) 4/13 x 5/12 + 4/13 x 7/ 12 - 4/3

= 4/13 x ( 5/12 + 7/12 - 13/3 )

= 4/13 x ( - 10/3 )

= -40/39 

a) \(\frac{9}{13}\cdot\frac{7}{12}+\frac{9}{13}\cdot\frac{5}{12}-\frac{9}{13}\)

\(=\frac{9}{13}.\left(\frac{7}{12}+\frac{5}{12}-1\right)=\frac{9}{13}.\left(1-1\right)=\frac{9}{13}.0=0\)

b) \(\frac{4}{13}\cdot\frac{5}{12}+\frac{4}{13}\cdot\frac{7}{12}-\frac{4}{3}\)

\(=\frac{4}{13}.\left(\frac{5}{12}+\frac{7}{12}\right)-\frac{4}{3}=\frac{4}{13}.1-\frac{4}{3}=\frac{4}{13}-\frac{4}{3}\)

1) = \(\frac{3}{5}\)

2) =\(\frac{6}{7}\)

3)\(\frac{9}{13}\)

4)\(\frac{4}{13}\)