a) Để \(-1:x\)là số nguyên 

\(\Rightarrow\)\(x\inƯ\left(-1\right)\in\left\{\pm1\right\}\)

Vậy \(x\in\left\{-1;1\right\}\)

b) Để \(1:x+1\)là số nguyên 

\(\Rightarrow\)\(x+1\inƯ\left(1\right)\in\left\{\pm1\right\}\)

+ \(x+1=1\)\(\Leftrightarrow\)\(x=1-1=0 \left(TM\right)\)

+ \(x+1=-1\)\(\Leftrightarrow\)\(x=-1-1=-2\left(TM\right)\)

Vậy \(x\in\left\{-2; 0\right\}\)

c) Để \(-2:x\)là số nguyên 

\(\Rightarrow\)\(x\inƯ\left(-2\right)\in\left\{\pm1;\pm2\right\}\)

Vậy \(x\in\left\{-1;-2;1;2\right\}\)

d) Để \(3:x-2\)là số nguyên 

\(\Rightarrow\)\(x-2\inƯ\left(3\right)\in\left\{\pm1;\pm3\right\}\)

- Ta có bảng giá trị:

Vậy \(x\in\left\{-1;1;3;5\right\}\)

e) Ta có: \(x+8=\left(x-7\right)+15\)

- Để \(x+8⋮x-7\)\(\Leftrightarrow\)\(\left(x-7\right)+15⋮x-7\)mà \(x-7⋮x-7\)

\(\Rightarrow\)\(15⋮x-7\)\(\Rightarrow\)\(x-7\in\left\{\pm1;\pm3;\pm5;\pm15\right\}\)

- Ta có bảng giá trị:

Vậy \(x\in\left\{-8;2;4;6;8;10;12;22\right\}\)

f) Ta có: \(2x+9=\left(2x-10\right)+19=2.\left(x-5\right)+19\)

- Để \(2x+9⋮x-5\)\(\Leftrightarrow\)\(2.\left(x-5\right)+19⋮x-5\)mà \(2.\left(x-5\right)⋮x-5\)

\(\Rightarrow\)\(19⋮x-5\)\(\Rightarrow\)\(x-5\inƯ\left(19\right)\in\left\{\pm1;\pm19\right\}\)

- Ta có bảng giá trị:

Vậy \(x\in\left\{-14;4;6;24\right\}\)

g) Ta có: \(2x+16=\left(2x-16\right)+32=2.\left(x-8\right)+32\)

- Để \(2x+16⋮x-8\)\(\Leftrightarrow\)\(2.\left(x-8\right)+32⋮x-8\)mà \(2.\left(x-8\right)⋮x-8\)

\(\Rightarrow\)\(32⋮x-8\)\(\Rightarrow\)\(x-8\inƯ\left(32\right)\in\left\{\pm1;\pm2;\pm4;\pm8;\pm16;\pm32\right\}\)

- Ta có bảng giá trị:

Vậy \(x\in\left\{-24;-8;0;4;6;7;9;10;12;16;24;40\right\}\)

h) Ta có: \(5x+2=\left(5x-5\right)+7=5.\left(x-1\right)+7\)

- Để \(5x+2⋮x-1\)\(\Leftrightarrow\)\(5.\left(x-1\right)+7⋮x-1\)mà \(5.\left(x-1\right)⋮x-1\)

\(\Rightarrow\)\(7⋮x-1\)\(\Rightarrow\)\(x-1\inƯ\left(7\right)\in\left\{\pm1;\pm7\right\}\)

- Ta có bảng giá trị:

Vậy \(x\in\left\{-6;0;2;8\right\}\)

k) Ta có: \(3x=\left(3x-6\right)+6=3.\left(x-2\right)+6\)

- Để \(3x⋮x-2\)\(\Leftrightarrow\)\(3.\left(x-2\right)+6⋮x-2\)mà \(3.\left(x-2\right)⋮x-2\)

\(\Rightarrow\)\(6⋮x-2\)\(\Rightarrow\)\(x-2\inƯ\left(6\right)\in\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

- Ta có bảng giá trị:

Vậy \(x\in\left\{-4;-1;0;1;3;4;5;8\right\}\)

bai nay kho qua

k minh minh giai cho

c,x-1 là ước của 5 

\(\Rightarrow5⋮x-1\Rightarrow x-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow x\in\left\{2;0;6;-4\right\}\)

Vậy.......................

d,\(7⋮3x+2\)

\(\Rightarrow3x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x\in\left\{-\frac{1}{3};-1;\frac{5}{3};-3\right\}\)

Vậy.........................

e;\(x+2⋮x-1\Rightarrow\left(x-1\right)+3⋮x-1\)

\(\Rightarrow3⋮x-1\Rightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow x\in\left\{2;0;4;-2\right\}\)

Vậy..........................

f;\(2x+1⋮x-3\Rightarrow2\left(x-3\right)+7⋮x-3\)

\(\Rightarrow7⋮x-3\Rightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow4;2;10;-4\)

Vậy.............................

g,\(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+......+\left(x-99\right)+\left(x-100\right)=-5750\)

\(\Rightarrow\left(x+x+x+.....+x+x\right)-\left(1+2+3+......+99+100\right)=-5750\)

\(\Rightarrow100x-5050=-5750\)

\(\Rightarrow100x=-700\)

\(\Rightarrow x=-7\)

Bài 1: 

a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{2;0;4;-2\right\}\)

(a) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{x^2-5x+9}{x-3}\in Z\)

Ta có: \(\dfrac{x^2-5x+9}{x-3}\left(x\ne3\right)=\dfrac{x\left(x-3\right)-2\left(x-3\right)+3}{x-3}=x-2+\dfrac{3}{x-3}\)nguyên khi và chỉ khi: \(\left(x-3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\\x-3=3\\x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\\x=6\\x=0\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{0;2;4;6\right\}\).

(b) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{2x^3-x^2+6x+2}{2x-1}\in Z\left(x\ne\dfrac{1}{2}\right)\)

Ta có: \(\dfrac{2x^3-x^2+6x+2}{2x-1}=\dfrac{x^2\left(2x-1\right)+3\left(2x-1\right)+5}{2x-1}=x^2+3+\dfrac{5}{2x-1}\)

nguyên khi và chỉ khi: \(\left(2x-1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=1\\2x-1=-1\\2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=3\\x=-2\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{-2;0;1;3\right\}\).

a: f(x) chia hết cho g(x)

=>x^2-3x-2x+6+3 chia hết cho x-3

=>3 chia hết cho x-3

=>x-3 thuộc {1;-1;3;-3}

=>x thuộc {4;2;6;0}

b: f(x) chia hết cho g(x)

=>2x^3-x^2+6x-3+5 chia hết cho 2x-1

=>5 chia hết cho 2x-1

=>2x-1 thuộc {1;-1;5;-5}

=>x thuộc {2;0;3;-2}