a) để B là phân số

=> 2x-1\(\ne\)0

=>2x\(\ne\)1

=>x\(\ne\)\(\frac{1}{2}\)

b) sửa đề :Tìm x để B có giá trị là  1 số nguyên

để B nguyên => x\(\in\)Z

=> 2x+5\(⋮\)2x-1

ta có : 2x-1\(⋮\)2x-1

=>(2x-5)-(2x-1)\(⋮\)2x-1

=>-4\(⋮\)2x-1

=>2x-1\(\in\)Ư(-4)={\(\pm1;\pm2;\pm4\)}

ta có bảng :

Mà x \(\in Z\)

nên x\(\in\){1;0}

1: Để A nguyên thì x+3-4 chia hết cho x+3

=>\(x+3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;1;-7\right\}\)

2: Để B nguyên thì 2x+4-9 chia hết cho x+2

=>\(x+2\in\left\{1;-1;3;-3;9;-9\right\}\)

=>\(x\in\left\{-1;-3;1;-5;7;-11\right\}\)

ĐKXĐ: \(x\ne\pm3\)

a

Khi x = 1:

\(A=\dfrac{3.1+2}{1-3}=\dfrac{5}{-2}=-2,5\)

Khi x = 2:

\(A=\dfrac{3.2+2}{2-3}=-8\)

Khi x = \(\dfrac{5}{2}:\)

\(A=\dfrac{3.2,5+2}{2,5-3}=\dfrac{9,5}{-0,5}=-19\)

b

Để A nguyên => \(\dfrac{3x+2}{x-3}\) nguyên

\(\Leftrightarrow3x+2⋮\left(x-3\right)\\3\left(x-3\right)+11⋮\left(x-3\right) \)

Vì \(3\left(x-3\right)⋮\left(x-3\right)\) nên \(11⋮\left(x-3\right)\)

\(\Rightarrow\left(x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\\ \Rightarrow x\left\{4;2;-8;14\right\}\)

c

Để B nguyên => \(\dfrac{x^2+3x-7}{x+3}\) nguyên

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

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

\(\Rightarrow x=\left\{-4;-11;-2;4\right\}\)

d

\(\left\{{}\begin{matrix}A.nguyên.\Leftrightarrow x=\left\{-8;2;4;14\right\}\\B.nguyên\Leftrightarrow x=\left\{-11;-4;-2;4\right\}\end{matrix}\right.\)

=> Để A, B cùng là số nguyên thì x = 4.