Question

HELPPPPPPPPPPP MEEE PLS

Answer 1

See explanation below

Explanation:

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

`\sqrt[n]{x}` =
`x^{(1)/(n)` Rule 1

`(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)^(1)/(3) (y^5)^(1)/(3) = x^(2)/(3)} . y ^ (5)/(3)` Using rules 1 and 3

The denominator is

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

Therefore the original expression in exponent form is

`(x^(2)/(3) y ^ (5)/(3))/(x^(3)/(4) y ^ 1)}`

Using rule 2 we get the final expression as

`x^{(2)/(3)-(3)/(4)}$ $y^{(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

`$x^{-(1)/(12)}y^{(2)/(3)}$`

Proved