Question 1111718
{{{((x^2 - 5x -14)/(x^3 -6x^2-7x)*(x^2 -4x -5))/((x^2 +x -30)/(2x))}}}
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Is it like that?
Cancel an x in each DEN
= {{{((x^2 - 5x -14)/(x^2 -6x -7)*(x^2 -4x -5))/((x^2 +x -30)/2)}}}

Factor some
= {{{((x-7)*(x+2)/(x-7)*(x+1)*(x-5)/(x+1))/(((x+6)*(x-5))/2)}}}
Cancel in the top fraction:
= {{{((x+2)*(x-5)/1)/(((x+6)*(x-5))/2)}}}
Cancel x-5
= {{{((x+2)/1)/((x+6)/2)}}}
= {{{((2x+4)/2)/((x+6)/2)}}}
= {{{((2x+4))/((x+6))}}}
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It's easier to do these on paper.