a, Với x = 1 thì \(A=\frac{3x+2}{x-3}=\frac{3\cdot1+2}{1-3}=\frac{5}{-2}=\frac{-5}{2}\)

Với x = 2 thì \(A=\frac{3x+2}{x-3}=\frac{3\cdot2+2}{2-3}=\frac{8}{-1}=-\frac{8}{1}=-8\)

Với x =\(\frac{5}{2}\)thì : \(A=\frac{3x+2}{x-3}=\frac{3\cdot\frac{5}{2}+2}{\frac{5}{2}-3}=\frac{\frac{15}{2}+2}{\frac{5}{2}-3}=\frac{\frac{19}{2}}{-\frac{1}{2}}=\frac{19}{2}\cdot(-2)=\frac{19}{1}\cdot(-1)=-19\)

b, Ta có : \(\frac{3x+2}{x-3}=\frac{3x-9+11}{x-3}=\frac{3(x-3)+11}{x-3}=3+\frac{11}{x-3}\)

\(\Leftrightarrow11⋮x-3\Leftrightarrow x-3\inƯ(11)=\left\{\pm1;\pm11\right\}\)

Lập bảng :

c,Để suy nghĩ đã

Làm tiếp :v

c, \(B=\frac{x^2+3x-7}{x+3}=\frac{x(x+3)-7}{x+3}=x-\frac{7}{x+3}\)

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

Lập bảng :

d, Tương tự

a) ta có: \(A=\frac{2x}{x-2}=\frac{2x-4+4}{x-2}=\frac{2.\left(x-2\right)+4}{x-2}=\frac{2.\left(x-2\right)}{x-2}+\frac{4}{x-2}=2+\frac{4}{x-2}\)

Để \(A\inℤ\)

\(\Rightarrow\frac{4}{x-2}\inℤ\)

\(\Rightarrow4⋮x-2\Rightarrow x-2\inƯ_{\left(4\right)}=\left(4;-4;2;-2;1;-1\right)\)

nếu x -2 = 4 => x = 6 (TM)

x- 2= - 4 => x= - 2 (TM)

x- 2= 2 => x = 4 (TM)

x- 2 = -2 => x = 0 (TM)

x - 2 = 1 => x = 3 (TM) 

x - 2 = -1 => x=  1 (TM)

KL: \(x\in\left(6;-2;4;0;3;1\right)\)

c) ta có: \(C=\frac{x^2+2}{x+1}=\frac{\left(x+1\right).\left(x-1\right)+3}{x+1}=\frac{\left(x+1\right).\left(x-1\right)}{x+1}+\frac{3}{x+1}\)\(=x-1+\frac{3}{x+1}\)

Để \(C\inℤ\)

\(\Rightarrow\frac{3}{x+1}\inℤ\)

\(\Rightarrow3⋮x+1\Rightarrow x+1\inƯ_{\left(3\right)}=\left(3;-3;1;-1\right)\)

nếu x + 1 = 3 => x = 2 (TM)

x + 1 = - 3 => x = -4 (TM)

x + 1 = 1 => x = 0 

x + 1 = -1 => x = -2 (TM)

KL: \(x\in\left(2;-4;0;-2\right)\)

p/s

ẺGTGNHMJ

làm đi

a) \(A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-3x+2}\)

\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-x-2x+2}\)

\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x\left(x-1\right)-2\left(x-1\right)}\)

\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{\left(x-1\right)\left(x-2\right)}\)

\(\Leftrightarrow A=\frac{\left(4x-1\right)\left(x-1\right)-\left(x-3\right)\left(x-2\right)-2x+4}{\left(x-2\right)\left(x-1\right)}\)

\(\Leftrightarrow A=\frac{4x^2-4x-x+1-x^2+2x+3x-6-2x+4}{\left(x-2\right)\left(x-1\right)}\)

\(\Leftrightarrow A=\frac{3x^2-2x-1}{\left(x-2\right)\left(x-1\right)}\)

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

b)\(\frac{3x+1}{x-2}=\frac{3x-6+7}{x-2}=\frac{3x-6}{x-2}+\frac{7}{x-2}=3+\frac{7}{x-2}\)

Ta có : \(x-2\inƯ_7\left\{-7;-1;1;7\right\}\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-2=-7\\x-2=-1\\x-2=1\\x-2=7\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}\text{x=-5}\\\text{x=1}\\\text{x=3}\\\text{x}=9\end{array}\right.\)

\(\text{x}=1\) (loại)

Vậy giá trị nguyên tập hợp x là:

x=-5;3;9