a) \(A=\dfrac{x+4}{x-3}=\dfrac{x-3+7}{x-3}=\dfrac{x-3}{x-3}+\dfrac{7}{x-3}\)
\(=1+\dfrac{7}{x-3}\)
Để A \(\in Z\) \(\Leftrightarrow\dfrac{7}{x+3}\in Z\) \(\Leftrightarrow\left(x-3\right)\inƯ\left(7\right)\)
\(\Leftrightarrow\left(x-3\right)\in\left\{-7;-1;1;7\right\}\)
\(\Leftrightarrow x\in\left\{-4;2;4;10\right\}\)
b) \(B=\dfrac{4x^2-4x+10}{2x+1}=\dfrac{\left(4x^2-4x+3\right)+7}{2x+1}\)
\(=\dfrac{4x^2-4x+3}{2x+1}+\dfrac{7}{2x+1}\)\(=\left(2x-3\right)+\dfrac{7}{2x+1}\)
<=> Để B thuộc Z <=> \(\left(2x-3\right)+\dfrac{7}{2x+1}\) thuộc Z
<=> \(\dfrac{7}{2x+1}\in Z\) <=> \(\left(2x+1\right)\inƯ\left(7\right)\)
<=> \(\left(2x+1\right)\in\left\{-7;-1;1;7\right\}\)
<=> \(x\in\left\{-4;-1;0;3\right\}\) (t/m)
Vậy..................
