What is the inverse of (+z)/(\frac{z^2){z^2-z-1}} ?

The inverse of (+z)/(\frac{z^2){z^2-z-1}} is (1+z+sqrt(z^2+2z+5))/2 ,(1+z-sqrt(z^2+2z+5))/2

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inverse of (+z)/(\frac{z^2){z^2-z-1}}

inverse

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

\frac{1 + z + z ^{2} + 2 z + 5}{2}, \frac{1 + z - z ^{2} + 2 z + 5}{2}

Solution steps

y = \frac{z}{\frac{z ^{2}}{z ^{2} - z - 1}}

Replace

z

with

y

z = \frac{y}{\frac{y ^{2}}{y ^{2} - y - 1}}

Solve for

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

\frac{1 + z + z ^{2} + 2 z + 5}{2}, \frac{1 + z - z ^{2} + 2 z + 5}{2}

f (x) = \frac{x + 4}{2 x + 1}

f (x) = \frac{7 x}{5}

f (x) = \frac{x}{a x + 1}

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

lo g_{2} (x - 1) + 3

What is the inverse of (+z)/(\frac{z^2){z^2-z-1}} ?
The inverse of (+z)/(\frac{z^2){z^2-z-1}} is (1+z+sqrt(z^2+2z+5))/2 ,(1+z-sqrt(z^2+2z+5))/2