Question

giải giúp mình câu b) được không

Answer 1

\(A=\dfrac{x-3}{x+3}+\dfrac{3x}{x-3}+\dfrac{4x^2+9}{9-x^2}\left(x\ne\pm3\right)\)

\(=\dfrac{x^2-6x+9+3x^2+9x-4x^2-9}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{3x}{\left(x+3\right)\left(x-3\right)}\)

\(C=\dfrac{A}{B}=\dfrac{2x+6}{\left(x+3\right)\left(x-3\right)}=\dfrac{2}{x-3}\left(x\ne\pm3\right)\)

\(C\) nguyên \(\Leftrightarrow x-3\in\left\{2;-2;1;-1\right\}\)

\(\Leftrightarrow x\in\left\{5;1;4;2\right\}\) (TM)