Question

Решить неравенство x²- x ≥ 0

Answer 1

Ответ:

$x\in(-\infty;0]\cup[1;+\infty)$

Объяснение:

$x^2-x\geq 0;\\x(x-1)\geq 0;\\--[0]--[1]-- > _X\\~~~^+~~~~~~~^-~~~~~~~^+\\x\in(-\infty;0]\cup[1;+\infty)$

Answer 2

$\displaystyle x^2-x\geq 0\\x(x-1)\geq 0\\\\\left \{ {{x\geq 0} \atop {x-1\geq 0}} \right. \rightarrow \left \{ {{x\geq 0} \atop {x\geq 1}} \right. \rightarrow x \in [1, + \infty)\\\\\left \{ {{x\leq 0} \atop {x-1\leq 0}} \right. \rightarrow \left \{ {{x\leq 0} \atop {x\leq 1}} \right. \rightarrow x \in (- \infty,0]\\ \\x \in (- \infty,0] \cup [1, + \infty)$