Odpowiedź:
-5
Szczegółowe wyjaśnienie:
[tex]\sqrt2x-3<2x\qquad|-\sqrt2x\\\\-3<2x-\sqrt2x\\\\2x-\sqrt2x>-3\\\\(2-\sqrt2)x>-3\qquad|:(2-\sqrt2)>0\\\\x>-\dfrac{3}{2-\sqrt2}\cdot\dfrac{2+\sqrt2}{2+\sqrt2}\\\\x>-\dfrac{3\cdot(2+\sqrt2)}{2^2-(\sqrt2)^2}\\\\x>-\dfrac{3\cdot(2+\sqrt2)}{4-2}\\\\x>-\dfrac{3\cdot(2+\sqrt2)}{2}[/tex]
[tex]\sqrt2\approx1,41\\\\-\dfrac{3\cdot(2+\sqrt2)}{2}\approx-\dfrac{3\cdot(2+1,41)}{2}=-\dfrac{3\cdot3,41}{2}=-\dfrac{10,23}{2}=-5,115[/tex]
[tex]x>-5,115[/tex]
