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(a) \( \frac{\left(16 x^{2 / 3} y^{-2}\right)^{-3 / 2}}{2 x^{-2} y} \)

Ask by Santiago Norton. in South Africa
Jan 23,2025

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Answer

\[ \frac{x y^{2}}{128} \]

Solution

To simplify the expression: \[ \frac{\left(16 x^{2 / 3} y^{-2}\right)^{-3 / 2}}{2 x^{-2} y} \] let's break it down step by step. ### Step 1: Simplify the Numerator Start with the numerator: \[ \left(16 x^{2 / 3} y^{-2}\right)^{-3 / 2} \] Apply the exponent \(-\frac{3}{2}\) to each factor inside the parentheses: \[ 16^{-3/2} \cdot \left(x^{2/3}\right)^{-3/2} \cdot \left(y^{-2}\right)^{-3/2} \] Calculate each part: 1. **For 16\^{-3/2}:** \[ 16^{-3/2} = \left(16^{1/2}\right)^{-3} = 4^{-3} = \frac{1}{64} \] 2. **For \(x^{2/3}\) raised to \(-3/2\):** \[ \left(x^{2/3}\right)^{-3/2} = x^{(2/3) \cdot (-3/2)} = x^{-1} \] 3. **For \(y^{-2}\) raised to \(-3/2\):** \[ \left(y^{-2}\right)^{-3/2} = y^{-2 \cdot (-3/2)} = y^{3} \] Putting it all together: \[ \left(16 x^{2 / 3} y^{-2}\right)^{-3 / 2} = \frac{1}{64} \cdot x^{-1} \cdot y^{3} = \frac{y^{3}}{64x} \] ### Step 2: Simplify the Entire Expression Now, substitute the simplified numerator back into the original expression: \[ \frac{\frac{y^{3}}{64x}}{2 x^{-2} y} \] This can be rewritten as: \[ \frac{y^{3}}{64x} \div (2 x^{-2} y) = \frac{y^{3}}{64x} \cdot \frac{1}{2 x^{-2} y} \] Simplify the multiplication: \[ \frac{y^{3} \cdot 1}{64x \cdot 2 x^{-2} y} = \frac{y^{3}}{128 x \cdot x^{-2} y} \] Combine the \(x\) terms: \[ x \cdot x^{-2} = x^{-1} = \frac{1}{x} \] So, the expression becomes: \[ \frac{y^{3}}{128 \cdot \frac{1}{x} \cdot y} = \frac{y^{3} \cdot x}{128 y} = \frac{x y^{2}}{128} \] ### Final Simplified Form \[ \boxed{\dfrac{x y^{2}}{128}} \]

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Let's break down the expression step-by-step! First, simplify the numerator: \[ \left(16 x^{2/3} y^{-2}\right)^{-3/2} = 16^{-3/2} \left(x^{2/3}\right)^{-3/2} \left(y^{-2}\right)^{-3/2} \] Calculating the individual parts, we find: \[ 16^{-3/2} = \frac{1}{(16^{1/2})^3} = \frac{1}{(4)^3} = \frac{1}{64} \] For \( x^{2/3} \) raised to \(-3/2\): \[ \left(x^{2/3}\right)^{-3/2} = x^{(-2/3) \cdot (3/2)} = x^{-1} = \frac{1}{x} \] And for \( y^{-2} \) raised to \(-3/2\): \[ \left(y^{-2}\right)^{-3/2} = y^{(-2) \cdot (-3/2)} = y^{3} \] Putting this all together gives us: \[ \left(16 x^{2/3} y^{-2}\right)^{-3/2} = \frac{y^3}{64x} \] Now, tackle the denominator: \[ 2 x^{-2} y = 2 \cdot \frac{1}{x^{2}} \cdot y = \frac{2y}{x^{2}} \] Now combine the two parts: \[ \frac{\frac{y^3}{64x}}{\frac{2y}{x^{2}}} = \frac{y^3}{64x} \cdot \frac{x^{2}}{2y} = \frac{y^3 \cdot x^{2}}{128xy} \] Simplifying this overall expression: \[ = \frac{y^{3-1} \cdot x^{2-1}}{128} = \frac{xy^2}{128} \] So, the final simplified form of the expression is: \[ \boxed{\frac{xy^2}{128}} \]

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