Question

Решите неравенство x^2/2<2x+2/3

Answer 1

$\frac{ x^{2} }{2} <2x + \frac{2}{3}$
$x^{2} <4x+ \frac{4}{3}$
$x^{2} -4x- \frac{4}{3} <0$
$3x^{2} -12x- 4 <0$
D=144+48=192
$x1= \frac{12+ \sqrt{192} }{6} = \frac{4(3+ \sqrt{12} )}{6} = \frac{2(3+ \sqrt{12} )}{3}$
$x2= \frac{2(3- \sqrt{12} )}{3}$

+++++x2--------x1+++++++
Ответ:$( \frac{2(3- \sqrt{12}) }{3};\frac{2(3+ \sqrt{12}) }{3} )$

Answer 2

3x²-4x-4<0
D=16+48=64
x1=(4-8)/6=-2/3
x2=(4+8)/6=2
       +            _                +
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           -2/3            2
x∈(-2/3;2)