a) \(A=\dfrac{x+3}{x+2}=\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}\)

\(\Rightarrow5⋮x-2\Rightarrow x-2\inƯ\left(5\right)\)

\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\\x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\\x=7\\x=-3\end{matrix}\right.\)

b) \(B=\dfrac{1-2x}{x+3}=\dfrac{-2x+1}{x+3}\)

\(B\in Z\Rightarrow-2x+1⋮x+3\)

\(\Rightarrow-2x-6+7⋮x+3\)

\(\Rightarrow-2\left(x+3\right)+7⋮x+3\)

\(\Rightarrow7⋮x+3\)

\(\Rightarrow x+3\inƯ\left(7\right)\)

\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x+3=1\\x+3-1\\x+3=7\\x+3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\\x=4\\x=-10\end{matrix}\right.\)

\(A=\dfrac{x+3}{x-2}=\dfrac{x-2+5}{x-2}=1+\dfrac{5}{x-2}\)

Để \(A\in Z\) thì \(x-2\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

\(\Rightarrow x\in\left\{3;1;7;-3\right\}\)

Vậy \(x\in\left\{3;1;7;-3\right\}\) thì \(A\in Z\)

\(B=\dfrac{1-2x}{x+3}=\dfrac{-2x-6+7}{x+3}=\dfrac{-2\left(x+3\right)-7}{x+3}=-2+\dfrac{-7}{x+3}\)

Để \(B\in Z\) thì \(x+3\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x\in\left\{-2;-4;4;10\right\}\)

Vậy \(x\in\left\{-2;-4;4;10\right\}\) thì \(B\in Z\)

- Ta có: \(\frac{2x+3}{x+1}=\frac{\left(2x+2\right)+1}{x+1}=\frac{2.\left(x+1\right)+1}{x+1}\)( ĐKXĐ: \(x\ne-1\))

- Để \(a\inℤ\)\(\Leftrightarrow\)\(\frac{2x+3}{x+1}\inℤ\)\(\Leftrightarrow\)\(\frac{2.\left(x+1\right)+1}{x+1}\inℤ\)

- Để \(\frac{2.\left(x+1\right)+1}{x+1}\inℤ\)\(\Leftrightarrow\)\(2.\left(x+1\right)+1⋮x+1\)mà \(2.\left(x+1\right)⋮x+1\)

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

+ Với \(x+1=1\)                                       + Với \(x+1=-1\)   

   \(\Leftrightarrow x=0\left(TM\right)\)                                      \(\Leftrightarrow x=-2\)

Vậy \(x\in\left\{-2,0\right\}\)

Đk: x#-3 
Với (*), A= (- 2x - 6 + 7)/(x + 3) = -2 + 7/(x+3) 
A nguyên <=> x + 3 thuộc Ư(7)={1;-1;7;-7} 
=> S = {-2;- 4;4;-10}

Bài 1:

a: ĐKXĐ: x<>0; x<>3; x<>1

\(A=\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)

\(=\frac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}\)

\(=\frac{x^2-6x+9-x^2+9}{x\cdot\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6\left(x-3\right)}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6}{x}\cdot\frac{x}{2\left(x-1\right)}=\frac{-3}{x-1}\)

b: Để A là số nguyên thì -3⋮x-1

=>x-1∈{1;-1;3;-3}

=>x∈{2;0;4;-2}

Kết hợp ĐKXĐ, ta được: x∈{2;4;-2}

Bài 2:

a: \(x^3-2x^2=x^2\cdot x-x^2\cdot2=x^2\left(x-2\right)\)

b: \(y^2-2y-x^2+1\)

\(=\left(y^2-2y+1\right)-x^2\)

\(=\left(y-1\right)^2-x^2=\left(y-1-x\right)\left(y-1+x\right)\)

c: \(\left(x+1\right)^2-25\)

=(x+1+5)(x+1-5)

=(x-4)(x+6)

A=7/x+3 -2 để A thuộc Z thì x+3 là ước của 7.

=>x+3=(+1,-1;+7,-7)

x=-2 =>A=5                                x=4=>A=-1

x=-4=> A=-9                                x=-10=>A=-3

\(B=\frac{1-2x}{x+3}=\frac{1-2x-6+6}{x+3}=\frac{7-2\left(x+3\right)}{x+3}=\frac{7}{x+3}-2\)

Để \(A=\frac{7}{x+3}-2\) là số nguyên <=> \(\frac{7}{x+3}\) là số nguyên

=> x + 3 thuộc Ư(7) = { - 7; - 1; 1; 7 }

+ ) Với x + 3 = - 7 thì x = - 10 (TM)

+ ) Với x + 3 = - 1 thì x = - 4 (TM)

+ ) Với x + 3 = 1 thì x = - 2 (TM)

+ ) Với x + 3 = 7 thì x = 4 (TM)

Vậy x = { - 10; - 4; - 2; 4 }

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