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

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

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

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

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

\(\Rightarrow5⋮x-2\Rightarrow x-2\inƯ\left(5\right)\)

\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)

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

b) \(B=\dfrac{1-2x}{x+3}=\dfrac{-2x+1}{x+3}\)

\(B\in Z\Rightarrow-2x+1⋮x+3\)

\(\Rightarrow-2x-6+7⋮x+3\)

\(\Rightarrow-2\left(x+3\right)+7⋮x+3\)

\(\Rightarrow7⋮x+3\)

\(\Rightarrow x+3\inƯ\left(7\right)\)

\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x+3=1\\x+3-1\\x+3=7\\x+3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\\x=4\\x=-10\end{matrix}\right.\)

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

Để \(A\in Z\) thì \(x-2\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

\(\Rightarrow x\in\left\{3;1;7;-3\right\}\)

Vậy \(x\in\left\{3;1;7;-3\right\}\) thì \(A\in Z\)

\(B=\dfrac{1-2x}{x+3}=\dfrac{-2x-6+7}{x+3}=\dfrac{-2\left(x+3\right)-7}{x+3}=-2+\dfrac{-7}{x+3}\)

Để \(B\in Z\) thì \(x+3\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)

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

Vậy \(x\in\left\{-2;-4;4;10\right\}\) thì \(B\in Z\)

con ngu asi dan don tu chi 

cai con ngu si dan don

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: \(x\in\left\{2;-2;4\right\}\)