a: DKXĐ: \(x\notin\left\{3;-3\right\}\)

b: \(A=\left(\dfrac{x}{\left(x-3\right)\left(x+3\right)}+\dfrac{-1}{x-3}\right)\cdot\dfrac{x+3}{3}\)

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

c: Thay x=5 vào A, ta được:

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

d: Để A là số nguyên thì \(x-3\in\left\{1;-1\right\}\)

hay \(x\in\left\{4;2\right\}\)

ab, đk x khác 3 ; -3 

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

c, x^2 - 8x + 15 = 0 <=> (x-3)(x-5) = 0 <=> x = 3 (ktm) ; x= 5 

Thay x = 5 vào A ta được : A =-1/2 

d, \(\Rightarrow x-3\inƯ\left(-1\right)=\left\{\pm1\right\}\)

TH1 : x - 3 = 1 <=> x = 4 

TH2 : x - 3 = -1 <=> x = 2 

Ta có : A = \(\frac{x+2}{x-1}=\frac{x-1+3}{x-1}=1+\frac{3}{x-1}\)

Để A thuộc Z thì 3 chia hết cho x - 1

Hay x - 1 thuộc Ư(3) = {-3;-1;1;3}

Ta có bảng : 

Để A > 1 thì 3/x - 1 > 0

Do đó : x - 1 > 0 

=> x > 1 

Vậy x > 1 thì A > 1

a) ĐKXĐ: \(x\notin\left\{0;3;1\right\}\)

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

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

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

\(=\dfrac{-6x+18}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-3}{x-1}\)

b) Để A nguyên thì \(-3⋮x-1\)

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

\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;-2;4\right\}\)

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.

a: ĐKXĐ: x<>1; x<>-1

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

c: Để A nguyên thì x+1-2 chia hết cho x+1

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

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

ĐKXĐ : x + 2 \(\ne0\Rightarrow x\ne-2\)

Ta có A = \(\frac{1-x}{x+2}=\frac{-2-x+3}{x+2}=-1+\frac{3}{x+2}\)

Để A \(\inℤ\Rightarrow\frac{3}{x+2}\inℤ\Rightarrow3⋮x+2\Rightarrow x+2\inƯ\left(3\right)\)

=> \(x+2\in\left\{1;3;-1;-3\right\}\)

=> \(x\in\left\{-1;1;-3;-5\right\}\)

\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)