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The expression
[tex]\[ \frac{1}{x-1} - \frac{3}{x} + 1 \][/tex]
can be written as
[tex]\[ \frac{x^2 + bx + c}{x(x-1)} \][/tex]
Find the values of [tex]\( b \)[/tex] and [tex]\( c \)[/tex].



Answer :

GinnyAnswer
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]

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