Question 1127915
It is C
by putting the left side over a common denominator, one gets x^2-3x=1=2x-5 and x^2-5x+6=0
This factors into (x-3)(x-2), and the roots of that equation are 3 and 2.
But, when one substitutes two back into the equation, the denominator for one term on the left and the term on the right are 0, so they are undefined.
The only solution is x=3.