Question

X^2+2x-3>0 решите уравнение

Answer 1

$\displaystyle x^2+2x-3>0\\x^2+3x-x-3>0\\x(x+3)-(x+3)>0\\(x+3)(x-1)>0\\\\ \left \{ {{x+3>0} \atop {x-1>0}} \right.=\left \{ {{x>-3} \atop {x>1}} \right.=x \in(1, + \infty)\\\\ \left \{ {{x+3<0} \atop {x-1<0}} \right.=\left \{ {{x<-3} \atop {x<1}} \right. =x\in(- \infty,-3)\\\\ x \in (- \infty,-3)\cup (1,+ \infty)$