To solve the equation \(\frac{x}{x-1}-\frac{2x-5}{x^{2}-3x+2}=\frac{1}{x-2}\), we need to find a common denominator for the fractions on the left-hand side and then simplify the equation. The common denominator will be \((x-1)(x-2)\). After simplifying, we can solve for \(x\).
To solve the equation \(\frac{x}{x-1} - \frac{2x-5}{x^2 - 3x + 2} = \frac{1}{x-2}\), we first identify the common denominator for the fractions on the left-hand side. The common denominator is \((x-1)(x-2)\).
Rewrite each term with the common denominator:
\[
\frac{x(x-2)}{(x-1)(x-2)} - \frac{(2x-5)}{(x-1)(x-2)} = \frac{1}{x-2}
\]
Combine the fractions on the left-hand side:
\[
\frac{x(x-2) - (2x-5)}{(x-1)(x-2)} = \frac{1}{x-2}
\]
Simplify the numerator:
\[
x(x-2) - (2x-5) = x^2 - 2x - 2x + 5 = x^2 - 4x + 5
\]
So the equation becomes:
\[
\frac{x^2 - 4x + 5}{(x-1)(x-2)} = \frac{1}{x-2}
\]
Multiply both sides by \((x-1)(x-2)\) to eliminate the denominator:
\[
x^2 - 4x + 5 = x-1
\]
Rearrange the equation to form a standard quadratic equation:
\[
x^2 - 4x + 5 - x + 1 = 0 \implies x^2 - 5x + 6 = 0
\]
Solve the quadratic equation:
\[
x^2 - 5x + 6 = 0
\]
The solutions to this equation are:
\[
x = 3 \quad \text{and} \quad x = 2
\]
Check the solutions in the original equation:
- For \(x = 3\):
\[
\frac{3}{3-1} - \frac{2(3)-5}{3^2 - 3(3) + 2} = \frac{1}{3-2} \implies \frac{3}{2} - \frac{1}{2} = 1 \implies 1 = 1
\]
This is true.
- For \(x = 2\):
\[
\frac{2}{2-1} - \frac{2(2)-5}{2^2 - 3(2) + 2} = \frac{1}{2-2} \implies \frac{2}{1} - \frac{-1}{0} = \text{undefined}
\]
This is undefined because division by zero is not allowed.
Thus, the only valid solution is \(x = 3\).
Answered
