Ta có ;

| x | + | y | ≤ 3

=> | x | + | y | ≤ | 3 | mà | x | ≥ 0 ∀ x

=> | 0 | ≤ | x | ≤ | 3 |

Do x nguyên => | x | ∈ { | 0 | ; | 1 | ; | 2 | ; | 3 | }

=> x ∈ { 0 ; ±1 ; ±2 ; ±3 }

Xét các trường hợp :

+) | x | = | 0 | => | y | = 3 => y = ±3

+) | x | = 1 | => | y | = 2 => y = ±2

+) | x | = 2 => | y | = 1 => y = ±1

+) | x | = 3 => | y | = 0 => y = 0

Vậy ( x , y ) ∈ { ( 0 , 3 ) ; ( 0 , -3 ) ; ( 1 , 2 ) ; ( 1 , -2 ) ; ( 2 , 1 ) ; ( 2 , -1 ) : ( 3 , 0 ) ; ( -1 , 2 ) ; ( -1 , -2 ) ; ( -2 , 1 ) ; ( -2 , -1 ) ; ( -3 , 0 ) }

Ta có (-32)²⁷ = -32²⁷ = -(2⁵)²⁷ = -2^5.27 = -2¹³⁵ = - 2⁹⁶.2³⁹

(-18)³⁹ = -18³⁹ = -(2.9)³⁹ = -2³⁹.(3²)³⁹  = -2³⁹.3⁷⁸ 

Nhận thấy 2⁹⁶ = (2³)³² = 8³² < 9³² = (3²)³² = 3⁶⁴ < 3⁷⁸

=> 2⁹⁶ < 3⁷⁸

=> -2⁹⁶ > -3⁷⁸

=> - 2⁹⁶.2³⁹ > -2³⁹.3⁷⁸ 

=> (-32)²⁷ > (-18)³⁹

Ta có : 

A = | 2x + 1 | + | x + 1 | + | x + 2 |

=> A ≥ | 2x + 1 | + | x + 1 + x + 2 |

=> A ≥ | 2x + 1 | + | 2x + 3 |

=> A ≥ | -2x - 1 | + | 2x + 3 |

=> A ≥ | -2x - 1 + 2x + 3 |

=> A ≥ | 2 | = 2

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(-2x-1\right)\left(2x+3\right)\ge0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge-1\\x\le\frac{-1}{2}\end{cases}}\)\(\Leftrightarrow-1\le x\le\frac{-1}{2}\)

\(\frac{1}{2^{3x-1}}=\frac{1}{32}\)

\(\Rightarrow\frac{1}{2^{3x-1}}=\frac{1}{2^5}\)

=> 3x - 1 = 5

=> 3x = 6

=> x = 2