\(\displaystyle \frac{9^{\frac{2}{3}}\cdot 3^{\frac{5}{3}}}{6^{2}}=\frac{(3^{2})^{\frac{2}{3}}\cdot3^{\frac{5}{3}}}{6^{2}}=\frac{3^{\frac{4}{3}}\cdot 3^{\frac{5}{3}}}{6^{2}}\)

\(\displaystyle =\frac{3^\frac{9}{3}}{6^2}=\frac{3^3}{6^2}=\frac{27}{36}=\frac{3}{4}\).

Using the properties of exponents, we can either choose to subtract the exponents of the corresponding bases or rewrite the expression using negative exponents as such:

\(\displaystyle x^{3/2}x^{-3}y^{7/3}y^{-2/9}\)

Here, we combine the terms with corresponding bases by adding the exponents together to get

\(\displaystyle x^{-3/2}y^{19/9}\)

Placing the x term (since it has a negative exponent) in the denominator will result in the correct answer. It can be shown that simply subtracting the exponents of corresponding bases will result in the same answer.

\(\displaystyle \frac{y^{19/9}}{x^{3/2}}\)

We proceed as follows

\(\displaystyle 4^{2.5}\) 

Write \(\displaystyle 2.5\) as a fraction

\(\displaystyle 4^{5/2}\) 

The denominator of the fraction is a \(\displaystyle 2\), so it becomes a square root.

\(\displaystyle (^2\sqrt{4})^5\) 

Take the square root.

\(\displaystyle 2^5\) 

Raise to the \(\displaystyle 5th\) power.

\(\displaystyle 32\) 

Recall that when considering rational exponents, the denominator of the fraction tells us the "root" of the expression.

Thus in this case we are taking the fifth root of \(\displaystyle 32\).

The fifth root of \(\displaystyle 32\) is \(\displaystyle 2\), because \(\displaystyle 2\cdot2\cdot2\cdot2\cdot2 = 32\).

Thus, we have reduced our expression to \(\displaystyle 2^3 = 2\cdot2\cdot2 = 8\). 

\(\displaystyle \frac{2x^3y^4}{6x^4y}\)

Simplify the constants:

\(\displaystyle \frac{x^3y^4}{3x^4y}\)

Subtract the "x" exponents:

\(\displaystyle \frac{y^4}{3xy}\)

\(\displaystyle x^3-x^4=x^{3-4}=x^{-1}=\frac{1}{x}\) This is how the x moves to the denominator.

Finally subtract the "y" exponents:

\(\displaystyle \frac{y^3}{3x}\)

To remove the fractional exponents, raise both sides to the second power and simplify:

\(\displaystyle (3x^{\frac{3}{4}})^{2} = (x^{\frac{1}{2}})^{2}\)

\(\displaystyle 9x^{\frac{3}{2}} = x\)

\(\displaystyle \frac{9x^{\frac{3}{2}}}{x} = \frac{x}{x}\)

\(\displaystyle 9x^{\frac{1}{2}} = 1\)

Now solve for \(\displaystyle x\):

\(\displaystyle \frac{9x^{\frac{3}{2}}}{9} = \frac{1}{9}\)

\(\displaystyle x^{\frac{1}{2}} = \frac{1}{9}\)

\(\displaystyle (x^{\frac{1}{2}})^{2} = (\frac{1}{9})^{2}\)

\(\displaystyle x = \frac{1}{81}\)

To remove the rational exponent, cube both sides of the equation:

\(\displaystyle (x^{\frac{1}{3}})^{3} = (2)^{3}\)

Now simplify both sides of the equation:

\(\displaystyle x^{\frac{1}{3}\cdot 3} = 2^{3}\)

\(\displaystyle x^{1} = 2^{3}\)

\(\displaystyle x = 8\)

\(\displaystyle \frac{2a^3b^{28}c^2d^2e^4f}{3a^2b^{27}cd^3e^5f^2}\)

When dividing two exponents with the same base we subtract the exponents:

\(\displaystyle \frac{2abcd^{-1}e^{-1}f^{-1}}{3}\)

Negative exponents are dealt with based on the rule 

\(\displaystyle a^{-m}=\frac{1}{a^m}\):

\(\displaystyle \mathbf{\frac{2abc}{3def}}\)

When an exponent is raised to the power of another exponent, just multiply the exponents together.

\(\displaystyle y = (x^a)^b = x^{ab}\)

\(\displaystyle y = ((x+2)^{1/2})^2 = (x+2)^1\)

Subtract the "x" exponents and the "y" exponents vertically. Then add the exponents horizontally if they have the same base (subtract the "x" and subtract the "y" ones). Finally move the negative exponent to the denominator.

\(\displaystyle \frac{8xy^2}{2x^4y}*x^2y^3\rightarrow 4x^{-3}y*x^2y^3\rightarrow 4x^{-1}y^4=\frac{4y^4}{x}\)