Question

rozwiąz nierownosc 5x^2 -4 >0​

Answer 1

Ответ:

$(-\infty;-0,4\sqrt{5})\cup(0,4\sqrt{5};+\infty)$

Пошаговое объяснение:

$5x^2-4>0\\\\5(x^2-\frac{4}{5})>0\;|:5\\\\x^2-(\frac{2}{\sqrt{5}})^2>0\\\\x^2-(\frac{2\sqrt{5}}{5})^2>0\\\\x^2-(0,4\sqrt{5})^2>0\\\\(x-0,4\sqrt{5})(x+0,4\sqrt{5})>0$

///////////////////// -0,4√5 ________ 0,4√5 ////////////////////

$x\in(-\infty;-0,4\sqrt{5})\cup(0,4\sqrt{5};+\infty)$