[tex]a)\\\\4x^2-1\geq0\\x^2\geq\frac{1}{4}\\|x|\geq\frac{1}{2}\\x\geq\frac{1}{2} \vee x\leq-\frac{1}{2}\\x\in\langle\frac{1}{2}, +\infty)\bigcup(-\infty,-\frac{1}{2}\rangle\\\\b)\\4x^2-5x < x^2+4x\\3x^2-9x < 0\\x^2-3x < 0\\(x-\frac{3}{2})^2 < \frac{9}{4}\\|x-\frac{3}{2}| < \frac{3}{2}\\x < 3\wedge x > 0\\x\in(0,3)[/tex]