Solve the following:
$\displaystyle{3 \left[ \frac{2}{3} (2x-1) - (x+4) \right] \lt -(5 - 3x) + \frac{3}{2} (4x-2)}$
$\displaystyle{3 - 4(x+1) \le x - 3 \left[ \frac{1}{2} (2x-3) \right]}$
Solve the following inequalities
$\displaystyle{x^2-2x-3 \lt 0}$
$\displaystyle{-x^2-2x+3 \gt 0}$
$\displaystyle{-x^2-2x+3 \le 0}$
$\displaystyle{x^2+2x \le 3}$
$\displaystyle{x^2 \le 2x + 3}$
$\displaystyle{x^3-6x^2 \le 8-12x}$
$\displaystyle{x^3-x^2 \gt 3-3x}$
Determine which values of $x$ make the given expression positive, negative, and zero.
$\displaystyle{-2(x-5)^2}$
$\displaystyle{x^2-3x-40}$
$\displaystyle{x^3+3x^2+3x+1}$
$\displaystyle{x^2-7x+1}$
$\displaystyle{(x^2-4x+4) - (4x^3-2x^2+12x-8)}$
$\displaystyle{(x+1) (x-2)^2 (x+3) (x-5)^7}$
$\displaystyle{x^{2/3}(x-3)^{3/5}}$
$\displaystyle{(x-1)^{1/3}(x-2)^{2/3} + (x-1)^{4/3}(x-2)^{5/3}}$
$\displaystyle{\frac{(x+3)^2(5-x)^{\frac{3}{5}}}{(x^2-4)(x^2+x+1)}}$
$\displaystyle{\frac{(x-2)^3(x^2-9)^2}{x(x^2+6x+9)}}$
Solve the following:
$\displaystyle{\left| 5-3x \right| \gt 4}$
$\displaystyle{\left| 2 - 5x \right| \lt 3}$
$\displaystyle{\left| 7-2x \right| \lt 5}$
$\displaystyle{\left| \frac{4x-1}{2} \right| \ge 3}$
$\displaystyle{\left| 3x+4 \right| = 9}$
Solve $\displaystyle{\frac{(x-2)^{1/3}(2x+3)^2}{(x+5)^3(x^2+4)} \ge 0}$
For what values of $x$ do the following expressions represent real values?
$\displaystyle{\sqrt{x^2-5}}$
$\displaystyle{\sqrt{\frac{x-4}{x+4}}}$