2x + 27 = 9

2x = -18

x = -9

Vậy.........

\(x\in Z,3x+27=9\)

\(3x+27=9\)

\(\Rightarrow3x=-18\)

\(\Rightarrow x=-6\)

a) 311 - x + 82 = 46 + (x - 21)

=> 393 - x = 25  + x

=> 393 - 25 = x + x

=> 368 = 2x

=>x = 368 : 2

=> x = 184

b) -(x - 3 + 85) = (x + 70 - 71) - 5

=> -x + 3 - 85 = x + 70 - 71 - 5

=> -x - 82 = x - 6

=> -x -x = 6 + 82

=> -2x = 88

=> x = 88 : (-2)

=> x = -44

\(311-x+82=46+\left(x-21\right)\)

\(393-x=36+x-21\)

\(393-36+21=2x\)

\(x=139\)

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7x-10=2 mũ 2

7x-10=4

7x=14

x=2

7x - 10 = 2⁷ : 2⁵

=> 7x - 10 = 2²

=> 7x - 10 = 4

=> 7x = 14

=> x = 2

3+2.[x-2]=2^3-(-1)

=5.[x-2]=15

=15:5

=3

=>x=3+2=5

\(3+2\left|x-2\right|=2^3-\left(-1\right)\)

\(\Rightarrow3+2\left|x-2\right|=9\)

\(\Rightarrow2\left|x-2\right|=6\)

\(\left|x-2\right|=3\)

\(\hept{\begin{cases}x=5\\x=-1\end{cases}}\)

Dùng nguyên lí Dirichle bạn ạ

Số dư khi chia chia cho 4 chỉ có thể là một trong các số 0 ; 1 ; 2 ;3 

Nên trong 5 số bất kì đó phải tồn tại 2 số có cùng số dư khi chia cho 4

=> hiệu 2 số này chia hết cho 4

0-[-3]=3

0-[-7]=7

\(0-\left[-3\right]=0+3=3\)

\(0-\left[-7\right]=0+7=7\)

\(a,\)\(M=2+2^2+2^3+2^4+...+2^{2017}+2^{2018}\)

\(2M=2^2+2^3+2^4+2^5+....+2^{2018}+2^{2019}\)

\(M=2^{2019}-2\)

\(b,\)\(M=2+2^2+2^3+2^4+....+2^{2017}+2^{2018}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+....+\left(2^{2017}+2^{2018}\right)\)

\(=2\left(2+1\right)+2^3\left(2+1\right)+....+2^{2017}\left(2+1\right)\)

\(=3\left(2+2^3+...+2^{2017}\right)⋮3\)

\(x-36:18=12-15\)

\(\Rightarrow x-2=-3\)

\(\Rightarrow x=-1\)

x - 36 : 18 = 12 - 15

x - 2          = -3

x               = (-3) + 2

x               = -1

Vậy  x=-1 

\(2^{111}=\left(2^3\right)^{37}=8^{37}\)

\(3^{77}>3^{74}=\left(3^2\right)^{37}=9^{37}\)

Vì \(8^{37}< 9^{37}\Rightarrow2^{111}< 9^{37}< 3^{77}\)