ĐKXĐ: \(x\ne3\)

Với \(x\ne3\), ta có:

\(A=\dfrac{2x-5}{x-3}\) \(=\dfrac{2x-6+1}{x-3}\) \(=2+\dfrac{1}{x-3}\)

Để A nguyên thì \(\dfrac{1}{x-3}\) nguyên

                   \(\Leftrightarrow1⋮x-3\)

                   \(\Leftrightarrow x-3\inƯ\left(1\right)=\left\{1;-1\right\}\)

                   \(\Leftrightarrow x=\left\{4;2\right\}\)

Vậy với x ={4; 2} thì A là một số nguyên.

ĐKXĐ: \(x\ne3\)

Để A là một số nguyên thì \(2x-5⋮x-3\)

\(\Leftrightarrow2x-6+1⋮x-3\)

mà \(2x-6⋮x-3\)

nên \(1⋮x-3\)

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

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

hay \(x\in\left\{4;2\right\}\)(thỏa mãn ĐKXĐ)

Vậy: Để A nguyên thì \(x\in\left\{4;2\right\}\)

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 \(\in-1,-2,-5,-4\)

Lời giải:
Để $A$ nguyên thì $2x-3\vdots x+1$
$\Rightarrow 2(x+1)-5\vdots x+1$

$\Rightarrow 5\vdots x+1$

$\Rightarrow x+1\in \left\{1; -1; 5; -5\right\}$

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

\(D=\frac{2x+1}{x-3}=\frac{2x-6}{x-3}+\frac{7}{x-3}=2+\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

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

D= \(\frac{2x+1}{x-3}=2+\frac{7}{x-3}\)

để D dương thì x-3 là uocs của 7=(-1,1,-7,7)

xét từng TH: 

x-3=-1=> x=2

x-3=1=>x=4

x-3=-7=>x=-4

x-3=7=>x=10

các giá trị x là 2,4,-4,10

mấy bạn giúp mik với

giúp gì thế

\(y=\dfrac{2x-3}{x-2}=\dfrac{2\left(x-2\right)+1}{x-2}=2+\dfrac{1}{x-2}\in Z\\ \Leftrightarrow x-2\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{1;3\right\}\)