Solve the following inequality: $\frac{|x-1|}{x+2} < 1$.

Solve for $x$ the following inequality: $\frac{1}{|x|-3} \leq \frac{1}{2}$

The set of all real numbers $x$ for which ${x^2} - |x + 2| + x > 0$ is

The solution set of the inequation $\sqrt{x^2+x-2} > (1-x)$ is

Let $a$ and $b$ be two positive real numbers such that $a+2b \leq 1$. Let $A_1$ and $A_2$ be respectively the areas of circles with radii $ab^3$ and $b^2$. Then,the maximum possible value of $\frac{A_1}{A_2}$ is

If $\sqrt{9x^2+6x+1} < (2-x)$,then: