<s>[INST] Solve the inequality $\frac{3-x}{x^2-2x-3}\le\frac{3-x}{x^2+2x-3}$ Solve the inequality $$\dfrac{3-x}{x^2-2x-3}\le\dfrac{3-x}{x^2+2x-3}$$
We have $D: \begin{cases}x^2-2x-3\ne0\Rightarrow x\ne -1;3 \\ x^2+2x-3\ne0\Rightarrow x\ne-3;1\end{cases}$
Is the given equality equivalent (in $D$) to $$x^2-2x-3\ge x^2+2x-3\\\iff x\le0$$ So the solutions are $x\le0\cap D$? It seems like the inequalities I wrote aren't equivalent, because I don't get the answer.
 [/INST] $$\dfrac{3-x}{x^2-2x-3}\le\dfrac{3-x}{x^2+2x-3}$$
$$\implies(3-x)\Bigl(\dfrac1{x^2-2x-3}-\dfrac1{x^2+2x-3}\Bigr)\le0$$
$$\implies(3-x)\Bigl(\dfrac{x^2+2x-3-x^2+2x+3}{(x-3)(x+1)(x+3)(x-1)}\Bigr)\le0$$
$$\implies (3-x)\Bigl(\dfrac{4x}{(x-3)(x+1)(x+3)(x-1)}\Bigr)\le0$$
$$\implies \dfrac{4x}{(x+1)(x+3)(x-1)}\ge0$$
$$\implies x\in(\infty, -3)\cup(-1,0]\cup(1,3)\cup(3,\infty)$$
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