a: \(A=\dfrac{x^2-5x+6-x^2+x+2x^2-6}{x\left(x-3\right)}=\dfrac{2x^2-4x}{x\left(x-3\right)}=\dfrac{2x}{x-3}\)

a:

ĐKXĐ: x<>-1/2

Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì

\(2x^3+x^2+2x+1+1⋮2x+1\)

=>\(2x+1\inƯ\left(1\right)\)

=>2x+1 thuộc {1;-1}

=>x thuộc {0;-1}

b:

ĐKXĐ: x<>1/3

 \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)

=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1

=>2 chia hết cho 3x-1

=>3x-1 thuộc {1;-1;2;-2}

=>x thuộc {2/3;0;1;-1/3}

mà x nguyên

nên x thuộc {0;1}

c: 

ĐKXĐ: x<>2

\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)

=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)

=>\(x+2⋮x-2\)

=>x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;0;6;-2}

ĐKXĐ: \(x\notin\left\{-1;2;-2\right\}\)

a) Ta có: \(A=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)

\(=\left(\dfrac{\left(x+1\right)^2}{x^2-x+1}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)

\(=\left(\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{\left(x-2\right)\left(x+2\right)}{3x\left(x+2\right)}\)

\(=\dfrac{x^3+3x^2+3x+1-2x^2-4x+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x-2}{3x}\)

\(=\dfrac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{3x}{x-2}\)

\(=\dfrac{3x}{x-2}\)

b) Để A nguyên thì \(3x⋮x-2\)

\(\Leftrightarrow3x-6+6⋮x-2\)

mà \(3x-6⋮x-2\)

nên \(6⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(6\right)\)

\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

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

Kết hợp ĐKXĐ, ta được:

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

Vậy: Để A nguyên thì \(x\in\left\{3;1;4;0;5;8;-4\right\}\)

ĐKXĐ: \(x\ne1\)

Ta có: \(B=\dfrac{x^4-2x^3-3x^2+8x-1}{x^2-2x+1}\)

\(=\dfrac{x^4-2x^3+x^2-4x^2+8x-4+3}{x^2-2x+1}\)

\(=\dfrac{x^2\left(x^2-2x+1\right)-4\left(x^2-2x+1\right)+3}{x^2-2x+1}\)

\(=\dfrac{\left(x-1\right)^2\cdot\left(x^2-4\right)+3}{\left(x-1\right)^2}\)

\(=x^2-4+\dfrac{3}{\left(x-1\right)^2}\)

Để B nguyên thì \(3⋮\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-1\right)^2\inƯ\left(3\right)\)

\(\Leftrightarrow\left(x-1\right)^2\in\left\{1;3;-1;-3\right\}\)

mà \(\left(x-1\right)^2>0\forall x\) thỏa mãn ĐKXĐ

nên \(\left(x-1\right)^2\in\left\{1;3\right\}\)

\(\Leftrightarrow x-1\in\left\{1;9\right\}\)

hay \(x\in\left\{2;10\right\}\) (nhận)

Vậy: \(x\in\left\{2;10\right\}\)

Bài 1:

a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)

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

\(=\left(\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right)\cdot\dfrac{2x}{-\left(x-1\right)}\)

\(=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{-2x}{x-1}\)

\(=\left(-1\right)\cdot\dfrac{-2x}{x-1}=\dfrac{2x}{x-1}\)

b: Để P nguyên thì \(2x⋮x-1\)

=>\(2x-2+2⋮x-1\)

=>\(2⋮x-1\)

=>\(x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được:

\(x\in\left\{2;3\right\}\)

c: P<1

=>P-1<0

=>\(\dfrac{2x}{x-1}-1< 0\)

=>\(\dfrac{2x-x+1}{x-1}< 0\)

=>\(\dfrac{x+1}{x-1}< 0\)

=>-1<x<1

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}-1< x< 1\\x\ne0\end{matrix}\right.\)

a) \(P=\dfrac{2x+5}{x+3}\inℤ\left(x\inℤ;x\ne-3\right)\)

\(\Rightarrow2x+5⋮x+3\)

\(\Rightarrow2x+5-2\left(x+3\right)⋮x+3\)

\(\Rightarrow2x+5-2x-6⋮x+3\)

\(\Rightarrow-1⋮x+3\)

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

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

b) \(P=\dfrac{3x+4}{x+1}\inℤ\left(x\inℤ;x\ne-1\right)\)

\(\Rightarrow3x+4⋮x+1\)

\(\Rightarrow3x+4-3\left(x+1\right)⋮x+1\)

\(\Rightarrow3x+4-3x-3⋮x+1\)

\(\Rightarrow1⋮x+1\)

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

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

c) \(P=\dfrac{4x-1}{2x+3}\inℤ\left(x\inℤ;x\ne-\dfrac{3}{2}\right)\)

\(\Rightarrow4x-1⋮2x+3\)

\(\Rightarrow4x-1-2\left(2x+3\right)⋮2x+3\)

\(\Rightarrow4x-1-4x-6⋮2x+3\)

\(\Rightarrow-7⋮2x+3\)

\(\Rightarrow2x+3\in\left\{-1;1;-7;7\right\}\)

\(\Rightarrow x\in\left\{-2;-1;-5;2\right\}\)

a) P=\(\dfrac{2x+5}{x+3}=\dfrac{2\left(x+3\right)-2}{x+3}=\dfrac{2\left(x+3\right)}{x+3}-\dfrac{2}{x+3}=2-\dfrac{2}{x+3}\)

để \(P\inℤ\) thì \(\dfrac{2}{x+3}\inℤ\) hay 2 ⋮ (x-3) ⇒x+3 ϵ Ư2= (2,-2,1,-1)

ta có bảng sau:

Vậy x \(\in-1,-2,-5,-4\)

\(D=\dfrac{2x+4}{3x-1}\\ =>3D=\dfrac{6x+12}{3x-1}=\dfrac{2\left(3x-1\right)+14}{3x-1}=2+\dfrac{14}{3x-1}\)

Để 3D nguyên thì : \(\dfrac{14}{3x-1}\in Z\)

\(=>14⋮\left(3x-1\right)\\ =>3x-1\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

\(=>3x\in\left\{2;0;3;-1;8;-6;15;-13\right\}\\ =>x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3};\dfrac{8}{3};-2;5;-\dfrac{13}{3}\right\}\)

Mà x nguyên \(=>x\in\left\{0;1;-2;5\right\}\)

Do những giá trị trên chỉ là 3D nguyên nên chưa chắc D đã nguyên

Vậy thử lại thay từng giá trị x vào bt D

Kết luận : \(x\in\left\{0;1;-2;5\right\}\)