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%20%5C%7B%20%7B%7Bx%5E2%5Cgeq%204%7D%20%5Catop%20%7Bx-3%3C0%7D%7D%20%5Cright.%5C%5Cx%5E2%20%5Cgeq%204%3B%20%28x%5E2-4%29%5Cgeq%200%3B%20%28x-2%29%28x%2B2%29%5Cgeq%200%3B%20x%20%5Cin%20%28-oo%3B-2%5D%20%5Ccup%20%5B2%3B%2Boo%29%5C%5Cx-3%3C0%3B%20x%3C3 "\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")
найдем пересечение этих множеств
__/////////_ - 2 ____________ 2 __//////____ 3 __/////
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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%20%5C%7B%20%7B%7Bx%5E2%5Cgeq%204%7D%20%5Catop%20%7B%282-5x%29%5E2%3D%5Cgeq%2016%7D%7D%20%5Cright.%5C%5Cx%5E2-4%20%5Cgeq%200%3B%20%28x-2%29%28x%2B2%29%5Cgeq%200%3B%20x%20%5Cin%20%28-oo%3B-2%5D%20%5Ccup%20%5B2%3B%2Boo%29%5C%5C%282-5x%29%5E2-4%5E2%5Cgeq0%3B%20%282-5x-4%29%282-5x%2B4%29%20%5Cgeq0%3B%20%28-5x-2%29%28-5x%2B6%29%5Cgeq%200%5C%5Cx%20%5Cin%20%28-oo%3B%20-0.4%5D%20%5Ccup%20%5B1.2%3B%20%2Boo%29 "\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)")
найдем пересечение множеств
__/////_ - 2 _____ -0,4________1,2________2__////___
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Ответ : (-oo;-2]∪[2;+oo)