Đáp án:

Giải thích các bước giải:

 ` a) `

Để biểu thức có giá trị nguyên ` => x - 1 \in Ư( 5 ) = { -5 ; -1 ; 1 ; 5 } `

` => x \in { -4 ; 0 ; 2 ; 6 }`

` b) `

Để biểu thức có giá trị nguyên 

` => 2x + 7 \in 9 ={ -9 ; -3 ; -1 ; 1 ; 3 ; 9 } `

` => 2x \in { -16 ; -10 ; -8 ; -6 ; -4 ; 2 } `

` => x \in { -8 ; -5 ; -4 ; -2 ; 1 } `

` c) `

Để biểu thức có giá trị nguyên 

` => x + 4 \vdots x + 1 `

` => x + 1 + 3 \vdots x + 1 `

` => 3 \vdots x + 1 `

` => x + 1 \in Ư( 3 ) = { -3 ; -1 ; 1 ; 3 } `

` => x \in { -4 ; -2 ; 0 ; 2 } `

` d) `

Để biểu thức có giá trị nguyên

` => 10x - 8 \vdots x + 2 `

` => 10x + 20 - 28 \vdots x + 2 `

` => 10( x + 2 ) - 28 \vdots x + 2 `

` => 28 \vdots x + 2 `

` => x + 2 \in Ư( 28 ) = { -28 ; -14 ; -7 ; -4 ; -2 ; -1 ; 1 ; 2 ; 4 ; 7 ; 14 ; 28 }`

` => x \in { -30 ; -16 ; -9 ; -6 ; -4 ; -3 ; -1 ; 0 ; 2 ; 5 ; 12 ; 26 } `

`a,`

Với `x∈ZZ` để `A∈ZZ ⇔ 5:(x-1) ∈ZZ`                  đk `x \ne 1`

`⇒ x-1 ∈Ư(5) ={±1;±5}`

\begin{array}{|c|c|c|c|c|}\hline \text{$x-1$}&\text{$1$}&\text{$-1$}&\text{$5$}&\text{$-5$}\\\hline \text{$x$}&\text{$2(tm)$}&\text{$0(tm)$}&\text{$6(tm)$}&\text{$-4(tm)$}\\\hline\end{array}

Vậy `x∈{2;0;6;-4}` thì `A` nguyên

`b,`

Với `x∈ZZ` để `B∈ZZ ⇔ 9 :(2x+7) ∈ZZ`

`⇒ 2x+7 ∈Ư(9) ={±1;±3;±9}`

\begin{array}{|c|c|c|c|c|c|c|}\hline \text{$2x+7$}&\text{$1$}&\text{$-1$}&\text{$3$}&\text{$-3$}&\text{$9$}&\text{$-9$}\\\hline \text{$x$}&\text{$-3(tm)$}&\text{$-4(tm)$}&\text{$-2(tm)$}&\text{$-5(tm)$}&\text{$1(tm)$}&\text{$-8(tm)$}\\\hline\end{array}

Vậy `x∈{-3;-4;-2;-5;1;-8}` thì `C` nguyên

`c,`

Ta có `: (x+4) :(x+1) = 1 + 3 : (x+1)`                 đk `x \ne -1`

Với `x∈ZZ` để `C ∈ZZ ⇔ 3 : (x+1) ∈ZZ`

`⇒x+1 ∈Ư(3)={±1;±3}`

\begin{array}{|c|c|c|c|c|}\hline \text{$x+1$}&\text{$1$}&\text{$-1$}&\text{$3$}&\text{$-3$}\\\hline \text{$x$}&\text{$0(tm)$}&\text{$-2(tm)$}&\text{$2(tm)$}&\text{$-4(tm)$}\\\hline\end{array}

Vậy `x∈{0;2;-2;-4}` thì `C` nguyên

`d,`

Ta có `: (10x -8):(x+2) = 10 - 28 :(x+2)`               đk `x \ne -2`

Với `x∈ZZ` để `D∈ZZ ⇔ - 28 :(x+2) ∈ZZ`

`⇒ x+2 ∈Ư(-28) ={±1;±2;±4;±7;±14;±28}`

\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \text{$x+2$}&\text{$1$}&\text{$-1$}&\text{$2$}&\text{$-2$}&\text{$4$}&\text{$-4$}&\text{$7$}&\text{$-7$}&\text{$14$}&\text{$-14$}&\text{$28$}&\text{$-28$}\\\hline \text{$x$}&\text{$-1(tm)$}&\text{$-3(tm)$}&\text{$0(tm)$}&\text{$-4(tm)$}&\text{$2(tm)$}&\text{$-6(tm)$}&\text{$5(tm)$}&\text{$-9(tm)$}&\text{$12(tm)$}&\text{$-16(tm)$}&\text{$26(tm)$}&\text{$-30(tm)$}\\\hline\end{array}

Vậy `x∈{-1;-3;0;-4;2;-6;5;-9;12;-16;26;-30}` thì `D` nguyên