Question 989675
Pay most attention to the numerator.  THAT needs to be less than or equal zero.


{{{12x^2-4<=0}}}
{{{12x^2<=4}}}
{{{3x^2<=1}}}
{{{x^2<=1/3}}}
{{{x=0+- sqrt(1/3)}}}-----the critical value.



{{{3x^2-1}}}  is a parabola, less than 0 AT and BETWEEN the roots.
Roots found were {{{-sqrt(1/3)}}}  and {{{sqrt(1/3)}}};
Rationalized, {{{-sqrt(3)/3}}}  and  {{{sqrt(3)/3}}}.


ANSWER:
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Note that the denominator must not be zero, so {{{x<>1}}}.
Otherwise,  {{{-sqrt(3)/3<=x<=sqrt(3)/3}}} BUT  {{{x<>1}}} and {{{x<>-1}}}.
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