A: No.
Because (x-2) (2x-3)/(x-2) (3x+1) =1:
(x-2)(2x-3)=(x-2)(3x+ 1)
2x？ -7x+6=3x？ -5x-2
x？ +2x？ -8=0
(x+4)(x-2)=0
Solution: x1=-4; x2 = 2； Finally, the solution should be substituted into the original fractional equation for testing, and x2=2 (the denominator of the original equation is equal to 0) should be discarded.
X=-4 is the solution of the equation.
Although (2x-3)/(3x+1) =1;
2x-3=3x+ 1
x=-4
The final solution is also x =-4; But the solution to the problem is complete.
(2) When solving the equation x/(2x-3)=2x/(3x- 1), can we change the x on both sides of the equation into 1/(2x-3)=2/(3x- 1) to solve it? Why?
A: No.
Similarly, when x/(2x-3) = 2x/(3x-1);
x(3x- 1)=2x(2x-3)
3x？ -x=4x？ -6 times
x？ -5x=0
x(x-5)=0
Solution: x1= 0; x2 = 5； Finally, the solution should be substituted into the original fractional equation for testing, x1= 0; = 0; X2=5 is the solution of this equation.
However, when1/(2x-3) = 2/(3x-1);
3x- 1=2(2x-3)
3x- 1=4x-6
Solution: x = 5；; I omitted the solution of x=0.
To sum up, this is the topic of quadratic fractional equation, which can only be divided but not reduced; Otherwise, the root (solution) of the equation may be missed; In addition, the solution of the final fractional equation should be substituted into the original fractional equation for testing, and the solution that is not suitable for the problem should be discarded.