What is ((2x^3y^2z)/(3x^4yz^{-2)})^{-2} ?

The solution to ((2x^3y^2z)/(3x^4yz^{-2)})^{-2} is (9x^2)/(4y^2z^6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ((2x^3y^2z)/(3x^4yz^{-2)})^{-2}

simplify

(\frac{2 x ^{3} y ^{2} z}{3 x ^{4} y z ^{- 2}})^{- 2}

\frac{9 x ^{2}}{4 y ^{2} z ^{6}}

Solution steps

(\frac{2 x ^{3} y ^{2} z}{3 x ^{4} y z ^{- 2}})^{- 2}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

(\frac{2 x ^{3} y ^{2} z}{3 x ^{4} y z ^{- 2}})^{- 2} = \frac{1}{( \frac{2 x ^{3} y ^{2} z}{3 x ^{4} y z ^{- 2}} ) ^{2}}

= \frac{1}{( \frac{2 x ^{3} y ^{2} z}{3 x ^{4} y z ^{- 2}} ) ^{2}}

Simplify

(\frac{2 x ^{3} y ^{2} z}{3 x ^{4} y z ^{- 2}})^{2} : \frac{4 y ^{2} z ^{6}}{9 x ^{2}}

= \frac{1}{\frac{4 y ^{2} z ^{6}}{9 x ^{2}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{9 x ^{2}}{4 y ^{2} z ^{6}}

- 4 > x

\frac{\frac{9}{4}}{3}

10 v^{2} + 11 v - 8

4 ∣ x - 1 ∣ + 6 \leq 54

\frac{1 - \frac{3}{2}}{2}

What is ((2x^3y^2z)/(3x^4yz^{-2)})^{-2} ?
The solution to ((2x^3y^2z)/(3x^4yz^{-2)})^{-2} is (9x^2)/(4y^2z^6)