Question

Help solve pls fast
it's a maths question ​

Answer 1

Answer:

Step-by-step explanation:

$[(\sqrt[4]{x^{\dfrac{3}{4}}})^{\dfrac{-4}{3}} ]^{4}$

$= [(x^{\frac{3}{4}*\frac{1}{4}})^{\frac{-4}{3}}]^{4}\\\\\\= [(x^{\frac{3}{16} ) ^{\frac{-4}{3}}}]^{4}\\\\=[ x^{\frac{3}{16}*\frac{-4}{3}}]^{4}\\\\=[x^{\frac{-1}{4}}]^{4}\\\\=x^{\frac{-1}{4}*4}\\\\=x^{-1}\\\\=\dfrac{1}{x}$

Hint:

$(a^{m})^{n}=a^{m*n}$

Answer 2

Answer:

$1/x$

Step-by-step explanation:

To simplify, work from the inside out.

We start with

$\sqrt[4]{x^{3/4}}$

on the inside. And we can change the fourth root into a fractional exponent -- 1/4:

$(x^{\frac{3}{4})^{1/4}$

A power of a power means multiply the exponents, giving

$x^{\frac{3}{4} * \frac{1}{4}} = x^{3/16}$

So now we have

$[(x^{3/16})^{-4/3}]^4$

From here, apply the "power of a power rule" again, working from the inside out.

$(x^{-12/48})^4$

$x^{-48/48}$

$x^{-1}$

or

$1/x$