Question

Answer 1

Để A nguyên thì \(\Leftrightarrow x-3\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{4;2;14;-8\right\}\)

Answer 2

A=\(\dfrac{3x+2}{x-3}=\dfrac{3.\left(x-3\right)+11}{x-3}=3+\dfrac{11}{x-3}\)(x\(\ne3\))

để A nguyên<=> x-3 \(\in\)Ư₍₁₁₎={\(\pm11,\pm1\)}

=>\(\left[{}\begin{matrix}x-3=11\\x-3=-11\\x-3=1\\x-3=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=14\left(tm\right)\\x=-8\left(tm\right)\\x=4\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

vậy x\(\in\left\{14,4,2,-8\right\}\)thì A nguyên