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

Find the solution set of the inequality: $4x + 3 \geq 2x + 17$ and $3x - 5 < -2$.

Let $m, n$ be real numbers such that $0 \leq m \leq \sqrt{3}$ and $-\sqrt{3} \leq n \leq 0$. The minimum possible area of the region of the plane consisting of points $(x, y)$ satisfying the inequalities $y \geq 0$,$y - 3 \leq mx$,and $y - 3 \leq nx$ is

The set of all real values of $x$ satisfying the inequalities $x^2-4x+3 > 0$ and $x^2-2x-8 \leq 0$ 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

The solution set contained in $R$ of the inequation $3^x+3^{1-x}-4 < 0$ is: