\\ ==>\ x\in\left(-\infty;2-\sqrt3\right]\cup\left[2+\sqrt3;+\infty\right);" alt="(x-2)^2\geq3;\\ \left [ {{x-2\geq\sqrt3} \atop {x-2\leq-\sqrt3}} \right.\ \ \left [ {{x\geq2+\sqrt3} \atop {x\leq2-\sqrt3}} \right.:==>\\ ==>\ x\in\left(-\infty;2-\sqrt3\right]\cup\left[2+\sqrt3;+\infty\right);" align="absmiddle" class="latex-formula">
Имеем ответ:
x\geq2;==>x\in[2;+\infty);\\
x-2\geq9;\\
x\geq11\\
x\in[11;+\infty);" alt="\sqrt{(x-2)}\geq3;\\
D(f): x-2\geq0==>x\geq2;==>x\in[2;+\infty);\\
x-2\geq9;\\
x\geq11\\
x\in[11;+\infty);" align="absmiddle" class="latex-formula">