<s>[INST] solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$. 
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.

options
$a.)\ -101<x<25\\
b.)\ [-\infty,3]\\
c.)\ x\leq 3\\
\color{green}{d.)\ x<3}\\
$
I tried ,
Case $1$ ,for $ \boxed{x\geq 0}\\
\dfrac{x^2-x-12}{x-3}\geq 2x\\
\implies \dfrac{x^2-5x+12}{x-3}\leq 0 \\
\implies x<3\\
x\in \emptyset $
Case $2$ ,for $\boxed{x< 0}\\
\dfrac{x^2+x-12}{x-3}\geq 2x\\
\implies \dfrac{(x-4)(x-3)}{x-3}\leq 0 \\
\implies x\leq 4\\
\implies x< 0\\
  $
But the answer given is option $d.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.
 [/INST] Answer to question:
If we take $x=3$, we see that the equality doesn't hold, thus answers a, b and c are wrong. The only option left is d.
In case 1 you made an error, this equality doesn't hold:
$$\frac{x^2−5x+12}{x−3}≤0$$
 if you take $x=5$ we see that
$$\frac{5^2−5\cdot 5+12}{5−3} = \frac{25−25+12}{2} = \frac{12}{2}=6\geq 0$$.
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