What is 1/(n-1)+1/(n-3)-(n-1)/(n^2-4n+3) ?

The solution to 1/(n-1)+1/(n-3)-(n-1)/(n^2-4n+3) is 1/(n-1)

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simplify 1/(n-1)+1/(n-3)-(n-1)/(n^2-4n+3)

simplify

\frac{1}{n - 1} + \frac{1}{n - 3} - \frac{n - 1}{n ^{2} - 4 n + 3}

\frac{1}{n - 1}

Solution steps

\frac{1}{n - 1} + \frac{1}{n - 3} - \frac{n - 1}{n ^{2} - 4 n + 3}

Simplify

\frac{n - 1}{n ^{2} - 4 n + 3} : \frac{1}{n - 3}

= \frac{1}{n - 1} + \frac{1}{n - 3} - \frac{1}{n - 3}

Add similar elements:

\frac{1}{n - 3} - \frac{1}{n - 3} = 0

= \frac{1}{n - 1}

1272

(10 x^{3} - 4) (5 x^{6} - 2 x^{3} + 2)

3 \cdot \frac{2 3}{3}

(4 x + 3) (3 x - 8)

\frac{a \frac{2}{3}}{a \frac{3}{4}}

What is 1/(n-1)+1/(n-3)-(n-1)/(n^2-4n+3) ?
The solution to 1/(n-1)+1/(n-3)-(n-1)/(n^2-4n+3) is 1/(n-1)