Question

решить неравенство
[х-4]≤х

Answer 1

$|x-4| \leq x \\ \\ \sqrt{(x-4)^2} \leq x \ \ \ |^2 \\ \\ (x-4)^2 \leq x^2 \\ \\ x^2 -8x+16 \leq x^2 \\ \\ -8x+16 \leq 0 \\ \\ -8x \leq -16 \\ \\ x \geq 2$

Answer 2

$|x-4| \leq x\; \; \; \; \Rightarrow \; \; \; \; -x \leq x-4 \leq x\\\\ \left \{ {{x-4 \leq x} \atop {x-4 \geq -x}} \right. \; ,\; \left \{ {{-4 \leq 0} \atop {2x \geq 4}} \right. \; ,\; \left \{ {{x\in R} \atop {x \geq 2}} \right. \; \Rightarrow \; x \geq 2$