To solve the given problem, we need to rewrite the left-hand side expression [tex]\(\frac{1}{x-1} - \frac{3}{x} + 1\)[/tex] as a single rational expression with the same denominator as the right-hand side expression [tex]\(\frac{x^2 + bx + c}{x(x-1)}\)[/tex].
We start by finding a common denominator for the terms on the left-hand side:
[tex]\[
\frac{1}{x-1} - \frac{3}{x} + 1
\][/tex]
The common denominator for these fractions is [tex]\(x(x-1)\)[/tex].
Let's rewrite each fraction with this common denominator:
1. For [tex]\(\frac{1}{x-1}\)[/tex]:
[tex]\[
\frac{1}{x-1} = \frac{1 \cdot x}{(x-1) \cdot x} = \frac{x}{x(x-1)}
\][/tex]
2. For [tex]\(\frac{3}{x}\)[/tex]:
[tex]\[
\frac{3}{x} = \frac{3 \cdot (x-1)}{x \cdot (x-1)} = \frac{3(x-1)}{x(x-1)} = \frac{3x - 3}{x(x-1)}
\][/tex]
3. Let's rewrite the constant term 1:
[tex]\[
1 = \frac{x(x-1)}{x(x-1)} = \frac{x^2 - x}{x(x-1)}
\][/tex]
Now we combine all these terms:
[tex]\[
\frac{x}{x(x-1)} - \frac{3x - 3}{x(x-1)} + \frac{x^2 - x}{x(x-1)}
\][/tex]
Combine the numerators over the common denominator:
[tex]\[
\frac{x - (3x - 3) + (x^2 - x)}{x(x-1)}
\][/tex]
Simplify the numerator:
[tex]\[
x - 3x + 3 + x^2 - x = x^2 - 3x + 3 + 3
\][/tex]
[tex]\[
x^2 - 3x + 3
\][/tex]
So the left-hand side becomes:
[tex]\[
\frac{x^2 - 3x + 3}{x(x-1)}
\][/tex]
Now we need to match the numerator of this expression to the right-hand side form, [tex]\(x^2 + bx + c\)[/tex]:
[tex]\[
x^2 + bx + c = x^2 - 3x + 3
\][/tex]
By comparing coefficients, we see:
[tex]\[
b = -3, \quad c = 3
\][/tex]
Thus, the values of [tex]\(b\)[/tex] and [tex]\(c\)[/tex] are:
[tex]\[
b = -3, \quad c = 3
\][/tex]
