Question

Помогите пожалуйста. Решить неравенство. Во вложении.

Answer 1

ОДЗ
{(x+3)/(x-1)>0⇒x<-3 U x>1
{x>0
{x≠1
x∈(1;∞)⇒
(x+3)/(x-1)>x
(x+3)/(x-1) -x>0
(x+3-x²+x)/(x-1)>0
(-x²+2x+3)/(x-1)>0
(x²-2x-3)/(x-1)<0
x²-2x-3=0
x1+x2=2 U x1*x2=-3
x1=-1 U x2=3
x-1=0⇒x=1
                     \\\\\\\\\\\\\\\\\\\\\                       //////////////////////////
            _                     +                   _                 +
-----------------(-1)-------------(1)---------------(3)-----------------
                                            \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈(3;∞)