Respuesta :

WojtekR
Domain:

[tex]x-1\neq0\\\\\boxed{x\neq1}[/tex]

As we see in a numerator we have a square, so [tex](3x+1)^2[/tex] is always 0 or more. Now, there are two possibilities:

1. [tex](3x+1)^2>0[/tex]

We divide two numbers, and first one is positive, so the result will be also positive (it canot be 0 in this case) when [tex]x-1\ \textgreater \ 0[/tex]. So:

 [tex]x-1\ \textgreater \ 0\\\\\boxed{x\ \textgreater \ 1}[/tex]

2. [tex](3x+1)^2=0[/tex]

In this case the result wil be 0 (we have ≥ 0 in our inequality, so it will be a solution).

[tex](3x+1)^2=0\\\\3x+1=0\\\\3x=-1\qquad|:3\\\\\\\boxed{x=-\dfrac{1}{3}}[/tex]

So the solution to this inequality is:

[tex]\boxed{x=-\dfrac{1}{3} \qquad\vee\qquad x\ \textgreater \ 1}[/tex]
TobeySnaps

Alright! As you can see we got

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


So to do that we have to divide one or two numbers,

(3x+1)^2=0-3x+1=0


An also the first one is going to be positive.



Finally, the inequality is


x= -1/3

x > 1

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