1)
![\displaystyle\left \{ {{x^2\geq 4} \atop {x-3<0}} \right.\\x^2 \geq 4; (x^2-4)\geq 0; (x-2)(x+2)\geq 0; x \in (-oo;-2] \cup [2;+oo)\\x-3<0; x<3](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cleft+%5C%7B+%7B%7Bx%5E2%5Cgeq+4%7D+%5Catop+%7Bx-3%3C0%7D%7D+%5Cright.%5C%5Cx%5E2+%5Cgeq+4%3B+%28x%5E2-4%29%5Cgeq+0%3B+%28x-2%29%28x%2B2%29%5Cgeq+0%3B+x+%5Cin+%28-oo%3B-2%5D+%5Ccup+%5B2%3B%2Boo%29%5C%5Cx-3%3C0%3B+x%3C3)
найдем пересечение этих множеств
__/////////_ - 2 ____________ 2 __//////____ 3 __/////
\\\\\\\ \\\\\\\\ \\\\\\
смотрим где получилась "елочка"
Ответ (-oo;-2]∪[2;3)
2)
![\displaystyle\left \{ {{x^2\geq 4} \atop {(2-5x)^2=\geq 16}} \right.\\x^2-4 \geq 0; (x-2)(x+2)\geq 0; x \in (-oo;-2] \cup [2;+oo)\\(2-5x)^2-4^2\geq0; (2-5x-4)(2-5x+4) \geq0; (-5x-2)(-5x+6)\geq 0\\x \in (-oo; -0.4] \cup [1.2; +oo)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cleft+%5C%7B+%7B%7Bx%5E2%5Cgeq+4%7D+%5Catop+%7B%282-5x%29%5E2%3D%5Cgeq+16%7D%7D+%5Cright.%5C%5Cx%5E2-4+%5Cgeq+0%3B+%28x-2%29%28x%2B2%29%5Cgeq+0%3B+x+%5Cin+%28-oo%3B-2%5D+%5Ccup+%5B2%3B%2Boo%29%5C%5C%282-5x%29%5E2-4%5E2%5Cgeq0%3B+%282-5x-4%29%282-5x%2B4%29+%5Cgeq0%3B+%28-5x-2%29%28-5x%2B6%29%5Cgeq+0%5C%5Cx+%5Cin+%28-oo%3B+-0.4%5D+%5Ccup+%5B1.2%3B+%2Boo%29)
найдем пересечение множеств
__/////_ - 2 _____ -0,4________1,2________2__////___
\\\\\\\ \\\\\\ \\\\\\ \\\\\\
Ответ : (-oo;-2]∪[2;+oo)