` b) ` Để ` (2x - 1)/(x - 2) ∈ Z ` thì:
` (2x - 1) \vdots (x - 2) ` `(x ∈ Z, x ne 2)`
` => (2x - 4 + 3) \vdots (x - 2) `
` => [2(x - 2) + 3] \vdots (x - 2) `
Vì ` 2(x - 2) \vdots (x - 2) `
` => 3 \vdots (x - 2) `
` => (x - 2) ∈ Ư(3) = {±1 ; ±3} `
Ta có bảng giá trị:
\begin{array}{|c|c|c|}\hline x-2&-3&-1&1&3\\\hline x&-1&1&3&5\\\hline &tm&tm&tm&tm\\\hline \end{array}
Vậy ` x ∈ {-1 ; 1 ; 3 ; 5} ` thì ` (2x - 1)/(x - 2) ∈ Z `
` c) ` Để ` (3 - 5x)/(2x + 1) ∈ Z ` thì:
` (3 - 5x) \vdots 2x + 1 ` `(x ∈ Z, x ne -1/2)`
` => (-5x + 3) \vdots (2x + 1) `
` => (10x - 6) \vdots (2x + 1) `
` => (10x + 5 - 11) \vdots (2x + 1) `
Vì ` 10x + 5 = 5(2x + 1) \vdots (2x + 1) `
` => 11 \vdots (2x + 1) `
` => (2x + 1) ∈ Ư(11) = {±1 ; ±11} `
Ta có bảng giá trị:
\begin{array}{|c|c|c|}\hline 2x+1&-11&-1&1&11\\\hline 2x&-12&-2&0&10\\\hline x&-6&-1&0&5\\\hline &tm&tm&tm&tm\\\hline \end{array}
Vậy ` x ∈ {-6 ; -1 ; 0 ; 5} ` thì ` (3 - 5x)/(2x + 1) ∈ Z `
` d) ` Để ` (2x)/(1 - x) ∈ Z ` thì:
` 2x \vdots (1 - x) ` `(x ∈ Z, x ne 1)`
` => (2x - 2 + 2) \vdots (1 - x) `
Vì ` 2x - 2 = -2(1 - x) \vdots (1 - x) `
` => 2 \vdots (1 - x) `
` => (1 - x) ∈ Ư(2) = {±1 ; ±2} `
\begin{array}{|c|c|c|}\hline 1-x&-2&-1&1&2\\\hline x&3&2&0&-1\\\hline &tm&tm&tm&tm\\\hline \end{array}
Vậy ` x ∈ {3 ; 2 ; 0 ; -1} ` thì ` (2x)/(1 - x) ∈ Z `
