\(a.\dfrac{3x+1}{x^2+1}\ge0\)
Do : \(x^2+1>0\forall x\)
\(\Rightarrow3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)
KL ........
\(b.A=\dfrac{6x}{2x-1}=\dfrac{3\left(2x-1\right)+3}{2x-1}=3+\dfrac{3}{2x-1}\left(x\ne\dfrac{1}{2}\right)\)
Để : \(A\in Z\Leftrightarrow\dfrac{3}{2x-1}\in Z\Leftrightarrow2x-1\in\left\{\pm1;\pm3\right\}\)
\(\oplus2x-1=1\Leftrightarrow x=1\left(TM\right)\)
\(\oplus2x-1=-1\Leftrightarrow x=0\left(TM\right)\)
\(\oplus2x-1=3\Leftrightarrow x=2\left(TM\right)\)
\(\oplus2x-1=-3\Leftrightarrow x=-1\left(TM\right)\)
KL...........
\(c.B=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3+x}{3-x}-\dfrac{12x^2}{x^2-9}\right)=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{\left(3-x\right)\left(3+x\right)}{12x^2}=\dfrac{\left(x+1\right)\left(3+x\right)}{12x^3}\left(x\ne0;x\ne\pm3\right)\)
