Question 342197

a)

{{{(4x+2)^(-1) < 0}}}

{{{1/(4x+2) < 0}}}
<pre><b>
That is only true if the denominator is negative:

{{{4x+2 < 0}}}

{{{4x < -2}}}

{{{x<-2/4}}}

{{{x<-1/2}}}

--------------------
</pre></b>
b)
<pre><b>
{{{0 < (2x-4)^(-1) < 1/2}}}

{{{0 < 1/(2x-4) < 1/2}}}

We must requre that the denominator {{{(2x-4)}}} be positive,
so that we can multiple through by it without changing the order
of the inequality:

This requirement is 

{{{2x-4>0}}}
{{{2x>4}}}
{{{x>2}}}

{{{0 < 1/(2x-4) < 1/2}}}

{{{0 < 1/(2(x-2)) < 1/2}}}

With this requirement, we multiply through by the LCD = {{{2(x-2)}}}

{{{0 < 1 < x-2}}}

{{{1 < x-2}}}

{{{3 < x}}}

which is the same as 

{{{x > 3}}}

which is consistent with the requirement {{{x>2}}}

so the solution is 

{{{x>3}}}

Edwin</pre></b>