Question

Решите неравенство $\frac{x+1}{1-3x}\ \textgreater \ \frac{1}{3}$

Answer 1

x+1/1-3x-1/3>0

3(x+1)-(1-3x)/3(1-3x)>0

3x+3-1+3x/3(1-3x)>0

6x+2/3(1-3x)>0

2(3x+1)/3(1-3x)>0


{2(3x+1)>0
{3(1-3x)>0

{2(3x+1)<0
{3(1-3x)<0

{x>-1/3
{x<1/3

{x<-1/3
{x>1/3

x∈ (-1/3; 1/3)

Answer 2

$\frac{x+1}{1-3x}- \frac{1}{3}\ \textgreater \ 0\\\\ \frac{3x+3-1+3x}{3(1-3x)} \ \textgreater \ 0\\\\ \frac{6x+2}{3(1-3x)} \ \textgreater \ 0\\\\ \frac{6(x+ \frac{1}{3}) }{-9(x- \frac{1}{3}) } \ \textgreater \ 0\\\\(x+ \frac{1}{3})(x- \frac{1}{3})\ \textless \ 0$
         +                           -                             +
___________₀______________₀_______________
                  - 1/3                        1/3
x ∈ ( - 1/3 ; 1/3)