Example 1. Solve the equation: (x-1)/(x+2) = (3-x)/(2+x)+2.
Analysis: x- 1=3-x+2x+4 after removing the denominator.
Move items and merge items of the same category: 0x=8.
Because this equation has no solution, the original fractional equation has no solution.
Example 2. Solve the equation: (x2 +2)/( x2 -4)=2/(x+2)- 1.
Analysis: After removing the denominator: x2+2=2x-4-x2+4.
Move items and merge items of the same category: x2-x+ 1=0.
∫△= 1-4 & lt； ∴, this equation has no solution, ∴, the original equation has no solution.
Second, the fractional equation does not necessarily have no solution when generating additional roots.
If the integral equation transformed from the denominated fractional equation is a one-dimensional linear equation, its solution can make the simplest common denominator zero, and this root is an increasing root. Moreover, because a linear equation often has only one root, the original fractional equation has no solution at this time; If the transformed integral equation is a quadratic equation, the situation is different. Examples are as follows:
Example 3. Solve the equation:1(x-2)+3 = (1-x)/(2-x).
Analysis: The denominator is 1+3x-6=x- 1.
Solution: x=2
After testing, x=2 is an increasing root.
So the original equation has no solution.
Example 4. Solve the equation: x/(x-1)-2/(x+1) = 4/(x2-1).
Analysis: After removing the denominator: x2+x-2x+2=4.
Solution: x 1=2, x2=- 1.
It is verified that x=2 is the root of the original equation and x=- 1 is the added root.
So the root of the original equation is x=2.
Therefore, understanding the difference between increasing root and no solution can help us improve the correctness of solving fractional equation and has certain guiding significance for judging the solution of equation.