\( \sqrt [\frac{m}{n}]{ab}\cdot \sqrt [m]{ab}\cdot (\sqrt {ab})^{\frac{4m}{m}}\div (a^{\frac{1}{m}}\frac{}{}b^{\frac{2}{m}})^{2n}: \)

Let's dive into the expression you've provided: First, we can simplify each part of the expression separately. 1. The term \( \sqrt [\frac{m}{n}]{ab} = (ab)^{\frac{n}{m}} \). 2. The term \( \sqrt [m]{ab} = (ab)^{\frac{1}{m}} \). 3. The term \( (\sqrt {ab})^{\frac{4m}{m}} = (ab)^{\frac{1}{2} \cdot 4} = (ab)^{2} \). 4. The term \( (a^{\frac{1}{m}}b^{\frac{2}{m}})^{2n} = a^{\frac{2n}{m}}b^{\frac{4n}{m}} \). Now, combining all of these, we have: \[ (ab)^{\frac{n}{m}} \cdot (ab)^{\frac{1}{m}} \cdot (ab)^{2} \div (a^{\frac{2n}{m}}b^{\frac{4n}{m}}). \] Let's add the exponents for the terms on the numerator: - From \( (ab)^{\frac{n}{m}} \): exponent is \( \frac{n}{m} \). - From \( (ab)^{\frac{1}{m}} \): exponent is \( \frac{1}{m} \). - From \( (ab)^{2} \): exponent is \( 2 \). Combining these: \[ \frac{n}{m} + \frac{1}{m} + 2 = \frac{n + 1 + 2m}{m} = \frac{n + 1 + 2m}{m}. \] Now we also need to consider the denominator: - The exponent for \( a \) is \( -\frac{2n}{m} \) and for \( b \) is \( -\frac{4n}{m} \). Combining the fractions in the denominator results in: \[ \left( \frac{n + 1 + 2m}{m} \right) - \left(\frac{2n + 4n}{m}\right) = \frac{n + 1 + 2m - 6n}{m} = \frac{2m - 5n + 1}{m}. \] Hence, the final simplification gives you: \[ (ab)^{\frac{2m - 5n + 1}{m}}. \] So the final answer is: \[ (ab)^{\frac{2m - 5n + 1}{m}}. \]