1. a,n+13 chia hết cho n + 7 b, n-7 chia hết cho n+5 c, 2n + 13 chia hết cho n+5 d,5n +45 chia hết cho n+3

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1. Đáp án + Giải thích các bước giải:

   Bổ sung đề : Tìm `x∈Z`

   Ta có :

   `a,n+13=(n+7)+6`

   Vì `(n+7)` $\vdots$ `n+7`

   Nên để `n+13` $\vdots$ `n+7`

   Thì `6` $\vdots$ `n+7` `(ĐK:n+7\ne0→n\ne-7)`

   `→n+7∈Ư(6)`

   `→n+7∈{±1;±2;±3;±6}`

   `→n∈{-8;-6;-9;-5;-10;-4;-13;-1}` ( Thỏa Mãn )

   Vậy để `n+13` $\vdots$ `n+7` thì `n∈{-8;-6;-9;-5;-10;-4;-13;-1}`

   `——————-`

   `b,n-7=(n+5)-12`

   Vì `(n+5)` $\vdots$ `n+5`

   Nên để `n-7` $\vdots$ `n+5`

   Thì `12` $\vdots$ `n+5` `(ĐK:n+5\ne0→n\ne-5)`

   `→n+5∈Ư(12)`

   `→n+5∈{±1;±2;±3;±4;±6;±12}`

   `→n∈{-6;-4;-7;-3;-8;-2;-9;-1;-11;1;-17;7}` ( Thỏa Mãn )

   Vậy để `n-7` $\vdots$ `n+5` thì `n∈{-6;-4;-7;-3;-8;-2;-9;-1;-11;1;-17;7}`

   `———————`

   `c,2n+13=(2n+10)+3=2(n+5)+3`

   Vì `2(n+5)` $\vdots$ `n+5`

   Nên để `2n+13` $\vdots$ `n+5`

   Thì `3` $\vdots$ `n+5` `(ĐK:n+5\ne0→n\ne-5)`

   `→n+5∈Ư(3)`

   `→n+5∈{±1;±3}`

   `→n∈{-6;-4;-8;-2}` ( Thỏa Mãn )

   Vậy để `2n+13` $\vdots$ `n+5` thì `n∈{-6;-4;-8;-2}` 

   `——————-`

   `d,5n+45=(5n+15)+30=5(n+3)+30`

   Vì `5(n+3)` $\vdots$ `n+3`

   Nên để `5n+45` $\vdots$ `n+3`

   Thì `30` $\vdots$ `n+3` `(ĐK:n+3\ne0→n\ne-3)`

   `→n+3∈Ư(30)`

   `→n+3∈{±1;±2;±3;±5;±6;±10;±15;±30}`

   `→n∈{-4;-5;-6;-8;-9;-13;-18;-33;-2;-1;0;2;3;7;12;27}` ( Thỏa Mãn )

   Vậy để `5n+45` $\vdots$ `n+3` thì `n∈{-4;-5;-6;-8;-9;-13;-18;-33;-2;-1;0;2;3;7;12;27}`

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2. Đáp án:

    …

   Giải thích các bước giải:

    `a,n+13 \vdots n+7`

   `\to (n+7) +6 \vdots n+7`

   `\to 6 \vdots n+7`

   `\to n+7 ∈Ư(6)`

   `\to n+7 ∈{1,-1,2,-2,3,-3,6,-6}`

   `\to n∈ {-6,-8,-5,-9,-4,-10,-1,-13}`

   Vậy `n∈{-6,-8,-5,-9,-4,-10,-1,-13}`

   ,

   `b,n-7 \vdots n+5`

   `\to (n+5) -12 \vdots n+5`

   `\to -12 \vdots n+5`

   `\to n+5∈Ư(-12)`

   `\to n+5∈{1,-1,2,-2,3,-3,4,-4,6,-6,12,-12}`

   `\to n∈{-4,-6,-3,-7,-2,-8,-1,-9,1,-11,7,-17}`

   Vậy `n∈{-4,-6,-3,-7,-2,-8,-1,-9,1,-11,7,-17}`

   ,

   `c,2n+13 \vdots n+5`

   `\to 2n+10+3 \vdots n+5`

   `\to 2(n+5)+3 \vdots n+5`

   `\to 3 \vdots n+5`

   `\to n+7 ∈Ư(3)`

   `\to n+7∈{1,-1,-3,3}`

   `\to n∈{-6,-8,-10,-4}`

   Vậy `n∈{-6,-8,-10,-4}`

   ,

   `d,5n+45 \vdots n+3`

   `\to 5n+15+30 \vdots n+3`

   `\to 5(n+3)+30 \vdots n+3`

   `\to 30 \vdots n+3`

   `\to n+3 ∈Ư(30)`

   `\to n+3∈{1,-1,2,-2,3,-3,5,-5,6,-6,10,-10,15,-15,30,-30}`

   `\to n∈ {-2,-4,-1,-5,0,-6,2,-8,3,-9,7,-13,12,-18,17,-33}`

   Vậy `n∈ {-2,-4,-1,-5,0,-6,2,-8,3,-9,7,-13,12,-18,27,-33}`

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