What is the inverse of (4(x+1)^2)/((x-1)^2) ?

The inverse of (4(x+1)^2)/((x-1)^2) is (x+4+4sqrt(x))/(x-4),(x+4-4sqrt(x))/(x-4)

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inverse of (4(x+1)^2)/((x-1)^2)

inverse

\frac{4 ( x + 1 ) ^{2}}{( x - 1 ) ^{2}}

\frac{x + 4 + 4 x}{x - 4}, \frac{x + 4 - 4 x}{x - 4}

Solution steps

y = \frac{4 ( x + 1 ) ^{2}}{( x - 1 ) ^{2}}

Replace

x

with

y

x = \frac{4 ( y + 1 ) ^{2}}{( y - 1 ) ^{2}}

Solve for

y, x = \frac{4 ( y + 1 ) ^{2}}{( y - 1 ) ^{2}}

\frac{x + 4 + 4 x}{x - 4}, \frac{x + 4 - 4 x}{x - 4}

f (x) = (1)

y = 7 x - 5

\frac{1}{s ( s + \frac{1}{10} )}

x - 3 + 8

2 x^{2} + 3 x + 4

What is the inverse of (4(x+1)^2)/((x-1)^2) ?
The inverse of (4(x+1)^2)/((x-1)^2) is (x+4+4sqrt(x))/(x-4),(x+4-4sqrt(x))/(x-4)