as noted in prevous posts, we have to have x > 3, so one factor of the denominator > 0.
is the solution from your manual really the difference √(2x-1) - √(7x-21)? it seems that this should be a sum
(because as x-->4, √(2x-1) - √(7x-21) --> √7 - √7 = 0, which means the reciprocal would be unbounded).
again, it should be clear that if x > 3, x > 1, so √(2x-1) > 0, and √(7x - 21) > 0, so their sum is positive.
and i just realized something: in all my previous posts, i had this term as √(7x+21) (this stems from it being wrong in your post #8).
however, the arguments are still valid, because x > 3.5 > 3, so 7x-21 > 0, which is all my argument requires.
one thing about proofs of this nature is: watch your signs! i count at least 4 sign errors that have occured in this thread, 3 of which completely derail
your proof. anyway, continuing....
since both factors in your denominator are positive (by suitably keeping x "close enough to 4"), the absolute value delimiter can be removed.
that is, in our "shorthand" 5|a|/|b| = 5|a|/b, since b > 0.