[tex]\it c) 2x-1>x-5 \Rightarrow x>-4 \Rightarrow x\in(-4,\ \infty)\ \ \ \ (1)\\ \\ x-6\geq0 \Rightarrow x\geq6 \Rightarrow x\in[6,\ \infty)\ \ \ \ \ (2)\\ \\ (1).\ (2) \Rightarrow x\in(-4,\ \infty)\cap[6,\ \infty) \Rightarrow x\in[6,\infty)\ \ \ \ \ (3)[/tex]
[tex]\it x^2-3x-4=x^2+x-4x-4=x(x+1)-4(x+1)=(x+1)(x-4)[/tex]
Ultima inecuație devine:
[tex]\it (x+1)(x-4)\geq0 \Rightarrow x\in\mathbb{R}\backslash(-1,\ 4)\ \ \ \ (4)\\ \\ \\ (3),\ (4) \Rightarrow S=[6,\ \infty)[/tex]
