What is the inverse of (x^2-x+1)/(x^2-x) ?

The inverse of (x^2-x+1)/(x^2-x) is (1-x+sqrt(x^2+2x-3))/(2(1-x)),(1-x-sqrt(x^2+2x-3))/(2(1-x))

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inverse of (x^2-x+1)/(x^2-x)

inverse

\frac{x ^{2} - x + 1}{x ^{2} - x}

\frac{1 - x + x ^{2} + 2 x - 3}{2 ( 1 - x )}, \frac{1 - x - x ^{2} + 2 x - 3}{2 ( 1 - x )}

Solution steps

y = \frac{x ^{2} - x + 1}{x ^{2} - x}

Replace

x

with

y

x = \frac{y ^{2} - y + 1}{y ^{2} - y}

Solve for

y, x = \frac{y ^{2} - y + 1}{y ^{2} - y}

\frac{1 - x + x ^{2} + 2 x - 3}{2 ( 1 - x )}, \frac{1 - x - x ^{2} + 2 x - 3}{2 ( 1 - x )}

What is the inverse of (x^2-x+1)/(x^2-x) ?
The inverse of (x^2-x+1)/(x^2-x) is (1-x+sqrt(x^2+2x-3))/(2(1-x)),(1-x-sqrt(x^2+2x-3))/(2(1-x))