Đáp án:
`↓↓`
Giải thích các bước giải:
`c) ( n^2 - 5n ) ∈ B ( n - 2 )`
`=> n^2 - 5n vdots n - 2`
`=> n ( n - 2 ) - 3 ( n - 2 ) - 6 vdots n - 2`
`=> 6 vdots n - 2`
`=> n - 2 ∈ Ư(6) = { ±1 ; ±2; ±3; ±6 }`
`=> n - 2 ∈ { -1; 1; -2; 2; -3; 3; -6; 6 }`
`=> n ∈ { 1; 3; 0; 4; -1; 5; -4; 8}`
`d) ( 6n - 4 ) ∈ B ( 1 - 2n )`
`=> 6n - 4 vdots 1 - 2n`
`=> 6n - 4 vdots 2n - 1`
`=> 3 ( 2n - 1 ) -1 vdots 2n - 1`
`=> 1 vdots 2n - 1`
`=> 2n - 1 ∈ Ư(1) = ±1`
`=> 2n - 1 ∈ { -1; 1 }`
`=> 2n ∈ { 0; 2}`
`=> n ∈ { 0; 1 }`
`e) n + 1 ∈ Ư(n^2 +2n - 3 )`
`=> n^2 + 2n - 3 vdots n+1`
`=> n ( n + 1 ) + ( n + 1 ) - 4 vdots n+1`
`=> 4 vdots n+1`
`=> n + 1 ∈ Ư(4) = { ±1; ±2; ±4 }`
`=> n+1 ∈ { -1; 1; -2; 2; -4; 4 }`
`=> n ∈ { -2; 0; -3; 1; -5; 3 }`
