Question

(7x^2y-2)^-1/(4x^-1y)^-2

Answer 1

Answer:

$\boxed{\dfrac{16y^{4}}{7x^{4}}}$

Step-by-step explanation:

Your expression is

$\dfrac{(7x^{2}y^{-2})^{-1}}{(4x^{-1}y)^{-2}}$

1. Handle the outer exponents

Remember that x⁻¹ = 1/x . Then

$\dfrac{(7x^{2}y^{-2})^{-1}}{(4x^{-1}y)^{-2}} = \dfrac{(4x^{-1}y)^{2}}{7x^{2}y^{-2}}$

2. Square the numerator

When you square a number, you multiply its exponent by 2.

$\dfrac{(4x^{-1}y)^{2}}{7x^{2}y^{-2}} = \dfrac{4^{2}x^{-2}y^{2}}{7x^{2}y^{-2}}$

3. Divide like terms

$\dfrac{4^{2}x^{-2}y^{2}}{7x^{2}y^{-2}} = \left ( \dfrac{16}{7}\right)\left (\dfrac{x^{-2}}{x^{2}}\right )\left (\dfrac{y^{2}}{y^{-2}}\right )$

4. Evaluate each term

When you divide numbers with exponents, you subtract the exponent in the denominator from that in the numerator

$\left ( \dfrac{16}{7}\right)\left (\dfrac{x^{-2}}{x^{2}}\right )\left (\dfrac{y^{2}}{y^{-2}}\right ) = \left ( \dfrac{16}{7}\right)\left (\dfrac{x^{-4}}{1}\right )\left (\dfrac{y^{4}}{1}\right )$

5. Move x to the denominator and combine terms

$\left ( \dfrac{16}{7}\right)\left (\dfrac{x^{-4}}{1}\right )\left (\dfrac{y^{4}}{1}\right ) =\mathbf{ \dfrac{16y^{4}}{7x^{4}}}\\\\\text{The simplified expression with positive exponents is } \boxed{\mathbf{\dfrac{16y^{4}}{7x^{4}}}}$