ta thấy rằng 5 phải chia hết cho a tức là 

a(U)5=1,-1;5,-5

vậy a 1,-1,5,-5 thì x có giá trị nguyên 

\(M=\frac{a^4-16}{a^4-4a^3+8a^2-16a+16}=\frac{\left(a^2-4\right)\left(a^2+4\right)}{a^4-4a^3+4a^2+4a^2-16a+16}=\frac{\left(a-2\right)\left(a+2\right)\left(a^2+4\right)}{a^2\left(a^2-4a+4\right)+4\left(a^2-4a+4\right)}\)

\(=\frac{\left(a-2\right)\left(a+2\right)\left(a^2+4\right)}{\left(a^2+4\right)\left(a-2\right)^2}=\frac{a+2}{a-2}=\frac{a-2+4}{a-2}=1+\frac{4}{a-2}\)

Để \(M\in Z\Leftrightarrow a-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Ta có bảng:

Vậy...

\(a,\)\(A=\frac{a^2+4a+4}{a^3+2a^2-4a-8}\)

\(=\frac{\left(a+2\right)^2}{a^2\left(a+2\right)-4\left(a+2\right)}\)

\(=\frac{\left(a+2\right)^2}{\left(a+2\right)\left(a^2-4\right)}\)

\(=\frac{\left(a+2\right)^2}{\left(a+2\right)\left(a+2\right)\left(a-2\right)}\)

\(=\frac{1}{a-2}\)

\(a,A=\frac{\left(a+2\right)^2}{\left(a+2\right)\left(a^2-4\right)}=\frac{a+2}{\left(a-2\right)\left(a+2\right)}=\frac{1}{a-2}\)

b, Để  A có giá trị là một số nguyên thì \(1⋮a-2\)

=> \(\orbr{\begin{cases}a-2=1\\a-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3\\a=1\end{cases}}}\)

\(a,\)Để \(A\in Z\Rightarrow\frac{1}{x-2}\in Z\)\(\Rightarrow1\)\(⋮\)\(a-2\)

\(\Leftrightarrow a-2\inƯ_1\)

Mà \(Ư_1=\left\{1;-1\right\}\)

\(\Rightarrow\orbr{\begin{cases}a-2=1\\a-2=-1\end{cases}\Rightarrow\orbr{\begin{cases}a=3\\a=1\end{cases}}}\)

Vậy \(A\in Z\Leftrightarrow a\in\left\{1;3\right\}\)

a) \(ĐKXĐ:\hept{\begin{cases}a\ne\pm2\\a\ne1\\a\ne0\end{cases}}\)

\(A=\left(\frac{4a}{2+a}+\frac{8a^2}{4-a^2}\right):\left(\frac{a-3}{a^2-2a}-\frac{2}{a}\right)\)

\(\Leftrightarrow A=\frac{8a-4a^2+8a^2}{\left(2-a\right)\left(2+a\right)}:\frac{a-3-2a+4}{a\left(a-2\right)}\)

\(\Leftrightarrow A=\frac{4a^2+8a}{\left(2-a\right)\left(2+a\right)}:\frac{-a+1}{a\left(a-2\right)}\)

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

\(\Leftrightarrow A=\frac{4a^2\left(a-2\right)}{\left(a-2\right)\left(a-1\right)}\)

\(\Leftrightarrow A=\frac{4a^2}{a-1}\)

b) Để A nhận giá trị nguyên

\(\Leftrightarrow\frac{4a^2}{a-1}\inℤ\)

\(\Leftrightarrow4a^2⋮a-1\)

\(\Leftrightarrow4\left(a^2-1\right)+4⋮a-1\)

\(\Leftrightarrow4\left(a-1\right)\left(a+1\right)+4⋮a-1\)

\(\Leftrightarrow4⋮a-1\)

\(\Leftrightarrow a-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow a\in\left\{0;2;-1;3;-3;5\right\}\)

Ta sẽ loại các giá trị ở đkxđ

Vậy để \(A\inℤ\Leftrightarrow a\in\left\{2;-1;3;-3;5\right\}\)