Biến đổi \(A=4x^2+9x+29+\frac{4}{x-3}\)

\(\Leftrightarrow A\in Z\)

\(\Leftrightarrow\frac{4}{x-3}\in Z\)

\(\Leftrightarrow x-3\inƯ\left(4\right)\left\{\pm1;\pm2;\pm4\right\}\)

Ta có bảng :

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

\(=\frac{\left(x-2\right)\left(4x^2+3\right)}{\left(x-2\right)\left(2x+1\right)}=\frac{4x^2+3}{2x-1}\left(ĐKXĐ:x\ne2;x\ne-\frac{1}{2}\right)\)

b, \(A\in Z\Leftrightarrow\frac{4x^2+3}{2x-1}\in Z\Leftrightarrow2x+1+\frac{4}{2x-1}\in Z\)

\(\Leftrightarrow\frac{4}{2x-1}\in Z\Leftrightarrow4⋮\left(2x-1\right)\)

\(\Rightarrow2x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Mà 2x - 1 là số lẻ nên \(2x-1\in\left\{-1;1\right\}\Rightarrow x\in\left\{0;1\right\}\) (thỏa mãn ĐKXĐ)

\(M+\frac{2x^2}{\left(3-x\right)\left(x+1\right)}=\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{4x}{\left(3-x\right)\left(x+1\right)}\)
\(M=\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{4x}{\left(3-x\right)\left(x+1\right)}-\frac{2x^2}{\left(3-x\right)\left(x+1\right)}\)
\(M=\frac{2x\left(3-x\right)}{\left(3-x\right)\left(x-1\right)\text{​​}\left(x+1\right)}+\frac{4x\left(x-1\right)}{\left(3-x\right)\left(x-1\right)\left(x+1\right)}+\frac{2x^2\left(x-1\right)}{\left(3-x\right)\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{6x-2x^2+4x^2-4x+2x^3-2x^2}{\left(3-x\right)\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{2x^3-2x}{\left(3-x\right)\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{2x\left(x-1\right)}{\left(3-x\right)\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{2x}{\left(3-x\right)\left(x+1\right)}\)

có gì sai sót bạn bỏ qua
Học tốt 

b) Tìm điều kiện để M đc xác định
\(M=\frac{2x}{\left(3-x\right)\left(x+1\right)}\)
để M xác định thì 
3 - x ≠ 0  => x ≠ 3
x + 1 ≠ 0 => x ≠ -1
Vậy x ≠ { 3 ; -1 } thì M đc xác định
 

\(3-m=\frac{10}{x+2}\)

\(\Leftrightarrow\left(3-m\right)\left(x+2\right)=10\)

=> 3-m và x+2 thuộc Ư (10)={1;2;5;10}

TH1: \(\hept{\begin{cases}3-m=1\\x+2=10\end{cases}\Leftrightarrow\hept{\begin{cases}m=2\\x=8\end{cases}}}\)hoặc \(\hept{\begin{cases}3-m=10\\x+2=1\end{cases}\Leftrightarrow\hept{\begin{cases}m=-7\\x=1\end{cases}}}\)

TH2: \(\hept{\begin{cases}3-m=5\\x+2=2\end{cases}\Leftrightarrow\hept{\begin{cases}m=-2\\x=0\end{cases}}}\)hoặc \(\hept{\begin{cases}3-m=2\\x+2=5\end{cases}\Leftrightarrow\hept{\begin{cases}m=1\\x=-3\end{cases}}}\)(loại)

bài 3:

\(A=\frac{2x^3-6x^2+x-8}{x-3}\left(x\ne3\right)\)

\(\Leftrightarrow A=\frac{\left(2x^3-6x^2\right)+\left(x-8\right)}{x-3}=\frac{2x\left(x-3\right)+\left(x-8\right)}{x-3}=2x+\frac{x-8}{x-3}\)

Để A nguyên thì \(\frac{x-8}{x-3}\)nguyên 

Có: \(\frac{x-8}{x-3}=\frac{x-3-5}{x-3}=1-\frac{5}{x-3}\)

Vì x nguyên => x-3 nguyên => x-3 \(\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Ta có bảng