Solve $\displaystyle \frac{2x}{x^2-9}\le\frac{1}{x+2}$

------

$\displaystyle x^2-9\not=0$

.'. $\displaystyle x\in R-\{-3,3\}$

and

$\displaystyle x+2\not=0$

.'. $\displaystyle x\in R-\{-2\}$

then converting the original inequality to

$\displaystyle (2x)(x+2)\le(x^2-9)$

$\displaystyle (2x^2+4x)(-x^2+9)\le 0$

$\displaystyle (x^2+4x+9)\le 0$

As Discriminant <0 it has no real roots

so how to do further...

My assumptions:-

$\displaystyle x^2-9$should not be zero

$\displaystyle x+2$should also not be zero

Are my assumptions right?

Any help will be highly appreciated.