Answer:

Step-by-step explanation:
We have 
We are going to analyze both sides of the equation separately.
Part a: 
Applying common denominator:


Part b:

In the numerator we can apply common factor 4. And in the denominator we can apply difference of squares.
<em>Remember: </em><u><em>Difference of squares:</em></u><em>
</em>

Then,

An extraneous solution is <em>not a valid solution for a problem. </em>We know that the denominator can't be zero.
The denominator of the first side of the equation is:
and we have to see for which values of
the denominator is zero:
⇒
and 
The denominator of the second side of the equation is:

And we have to see for which values of
the expression is zero,
⇒
and 
Then the extraneous solutions of the equation are:

Because those are the values of
that make the denominator zero.