Question 597723
I believe that
{{{(k^2+k-42)/(1-k)}}}=(k^2+k-42)/(1-k) and {{{(k^2-k-30)/(k-1)}}}=(k^2-k-30)/(k-1) are your expressions for this problem.
When you cannot type a horizontal line sandwiched between polynomials, you need those parentheses.
 
A NOTE ABOUT ORDER OF OPERATIONS:
Without those parentheses, the meaning is different
k^2+k-42/1-k={{{k^2+k-42/1-k=k^2+k-42-k=k^2-42}}}
Otherwise, some people may understand what you meant. However, other people, along with all computers and calculators, will not know what you meant, and you will end up with the wrong result. I've seen that happen to chemists hastily calculating the results of an analysis.
 
THE DOMAIN is the same for those two rational expressions. Only one value of x (x=1) will make the denominator zero and thus is not allowed in the domain.
 
DIVIDING RATIONAL EXPRESSIONS
{{{(((k^2+k-42)/(1-k)))/(((k^2-k-30)/(k-1)))}}}={{{((k^2+k-42)/(1-k))}}}/{{{((k^2-k-30)/(k-1))}}}={{{((k^2+k-42)/(1-k))*((k-1)/(k^2-k-30))}}}={{{((k+7)(k-6)/(-1(k-1)))*((k-1)/(k-6)(k+5))}}}={{{(k+7)(k-6)(k-1)/((-1)(k-1)(k-6)(k+5))}}}
At this point, your teacher may expect you to do this:
{{{(k+7)(k-6)(k-1)/((-1)(k-1)(k-6)(k+5))=(k+7)cross((k-6))(k-1)/((-1)(k-1)cross((k-6))(k+5))=(k+7)cross((k-6))cross((k-1))/((-1)cross((k-1))cross((k-6))(k+5))=(k+7)/((-1)(k+5))=-((k+7)/(k+5))}}}
or something similar.
Some teacher may prefer
{{{(k+7)(k-6)(k-1)/((-1)(k-1)(k-6)(k+5))=(1/(-1))((k+7)/(k+5))((k-1)/(k-1))((k-6)/(k-6))=(-1)((k+7)/(k+5))*1*1=(-1)((k+7)/(k+5))=-((k+7)/(k+5))}}}
or something similar.
Canceling all pairs of identical factors present in numerator and denominator of a rational expression yields an equivalent rational expression IN LOWEST TERMS.
 
COMMON DENOMINATOR AND ADDITION:
{{{(k^2+k-42)/(1-k)+(k^2-k-30)/(k-1)=(-1)(k^2+k-42)/((-1)(1-k))+(k^2-k-30)/(k-1)=(-k^2-k+42)/(k-1)+(k^2-k-30)/(k-1)=(-k^2-k+42+k^2-k-30)/(k-1)=(-2k+12)/(k-1)=-2(k+6)/(k-1)}}}
In the first step, numerator and denominator of the first rational expression were multiplied by (-1) to BUILD UP the expression so that both expressions would have the common denominator.
The terms {{{-k^2}}} and {{{k^2}}} are OPPOSITES and so they add to yield zero.