What is the inverse of (4x^2-1)/(5-7x^2) ?

The inverse of (4x^2-1)/(5-7x^2) is sqrt(-(-1-5x)/(4+7x)),-sqrt(-(-1-5x)/(4+7x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of (4x^2-1)/(5-7x^2)

inverse

\frac{4 x ^{2} - 1}{5 - 7 x ^{2}}

- \frac{- 1 - 5 x}{4 + 7 x}, - - \frac{- 1 - 5 x}{4 + 7 x}

Solution steps

y = \frac{4 x ^{2} - 1}{5 - 7 x ^{2}}

Replace

x

with

y

x = \frac{4 y ^{2} - 1}{5 - 7 y ^{2}}

Solve for

y, x = \frac{4 y ^{2} - 1}{5 - 7 y ^{2}}

- \frac{- 1 - 5 x}{4 + 7 x}, - - \frac{- 1 - 5 x}{4 + 7 x}

f (x) = - \frac{x - 2}{5}

f (x) = - \frac{x - 2}{6}

(t + 1)^{2}

3 - x + 9

What is the inverse of (4x^2-1)/(5-7x^2) ?
The inverse of (4x^2-1)/(5-7x^2) is sqrt(-(-1-5x)/(4+7x)),-sqrt(-(-1-5x)/(4+7x))