Question

HELPPPPPPPPPPP MEEE PLS

Answer 1

Answer:

See explanation below

Step-by-step explanation:

If the base of a radical exponent is the same then the following properties hole and x≠ 0

$\sqrt[n]{x}$ = $x^{\frac{1}{n}$   Rule 1

$\frac{x^m}{x^n} = x^{m-n}$   Rule 2

$(x^m)^n = x^{mn}$   Rule 3

The numerator in the expression is

$\sqrt[3]{x^2y^5} =(x^2)^\frac{1}{3} (y^5)^\frac{1}{3} = x^\frac{2}{3}} . y ^ \frac{5}{3}$  Using rules 1 and 3

The denominator is

$\sqrt[4]{x^3y^4} =(x^3)^\frac{1}{4} (y^4)^\frac{1}{4} = x^\frac{3}{4}} . y ^ 1}$ Using rules 1 and 3

Therefore the original expression in exponent form is

$\frac{x^\frac{2}{3} y ^ \frac{5}{3}}{x^\frac{3}{4} y ^ 1}}$  

Using rule 2 we get the final expression as

$x^{\frac{2}{3}-\frac{3}{4}} y^{\frac{5}{3}-1}$

2/3 - 3/4 = 8/12 - 9/12 = -1/2

5/3 - 1 = 5/3 - 3/3 = 2/3

So the final expression is

$x^{-\frac{1}{12}}y^{\frac{2}{3}}$

Proved