11A:
\(10n^2+n-10⋮n-1\)
=>\(10\left(n^2-1\right)+n-1+1⋮n-1\)
=>\(1⋮n-1\)
=>\(n-1\in\left\{1;-1\right\}\)
=>\(n\in\left\{2;0\right\}\)
11B;
\(2n^3-7n^2+13n+2⋮2n-1\)
=>\(2n^3-n^2-6n^2+3n+10n-5+7⋮2n-1\)
=>\(7⋮2n-1\)
=>\(2n-1\in\left\{1;-1;7;-7\right\}\)
=>\(2n\in\left\{2;0;8;-6\right\}\)
=>\(n\in\left\{1;0;4;-3\right\}\)
`10n^2 + n - 10 vdots n - 1 (n ne 1) `
`<=> 10n^2 - 10n + 10n - 10 + n - 1 + 1 vdots n - 1`
`<=> 10n(n-1) + 10(n-1) + (n-1) + vdots n - 1`
Do `n - 1 vdots n-1`
`=> 10n(n-1) + 10(n-1)+ (n-1) vdots n-1`
Khi đó: `1 vdots n-1`
`<=> n - 1 ∈ Ư(1) = {-1;1}`
`<=> n ∈ {0;2}` (Thỏa mãn)
Vậy ....
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`2n^3 - 7n^2 + 13n + 2 vdots 2n - 1 (n ne 1/2)`
`<=> 2n^3 - n^2 - 6n^2 + 3n + 10n - 5 + 7 vdots 2n - 1 `
`<=> (2n^3 - n^2) - (6n^2 - 3n) + (10n - 5) + 7 vdots 2n - 1`
`<=> n^2 (2n - 1) - 3n(2n - 1) + 5(2n - 1) + 7 vdots 2n - 1 `
Do `2n - 1 vdots 2n-1 `
`=> n^2 (2n - 1) - 3n(2n - 1) + 5(2n - 1) vdots 2n - 1`
Khi đó: `7 vdots 2n - 1`
`<=> 2n - 1 ∈ Ư(7) = {-7;-1;1;7}`
`<=> 2n ∈ {-6;0;2;8}`
`<=> n ∈ {-3;0;1;4}` (Thỏa mãn)
Vậy ...
