Zadanie 1
2(x+2)^2 - (\sqrt2x - 2)^2 \geq 0
2(x^2+4x+4)-(2x^2-4\sqrt2x+4)\geq0
2x^2+8x+8-2x^2+4\sqrt2x-4\geq 0
8x+4\sqrt2x+4\geq 0
4x(2+\sqrt2)\geq-4 |:4
x(2+\sqrt2)\geq -1
x\geq\frac{-1}{2+\sqrt2}
x\geq\frac{-(2-\sqrt2)}{(2+\sqrt2)(2-\sqrt2)}
x\geq \frac{\sqrt2-2}{4-2}
x\geq \frac{\sqrt2-2}{2}
x\in \langle\frac{\sqrt2-2}{2};+\infty)