Question

(1+х)²+3х²<(2х-1)²+7

Answer 1

$x^{2} +2x+1+3 x^{2} -4 x^{2} +4x-1-7<0$
$6x-7<0$
$6x<7$
$x<1 frac{1}{6}$

Ответ: $x$∈$(-$беск.$;1 frac{1}{6} )$