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6. \( \sqrt[3]{\frac{x^{2}}{y}} \cdot \sqrt[12]{\frac{27 x^{2}}{16 y}}: \sqrt[6]{\frac{9 x^{5}}{4 y^{2}}} \) 7. \( \sqrt[8]{\frac{x^{3}+x^{2} y}{x-y}}: \sqrt[4]{\frac{x^{2}+x y}{x-y}} \)

Ask by French Wheeler. in Italy
Jan 24,2025

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**Problema 6** \[ \frac{1}{3^{\frac{1}{12}} \, y^{\frac{5}{12}}} \] **Problema 7** \[ \sqrt[8]{\frac{x - y}{x + y}} \]

Solución

**Problema 6** Calcoliamo l'espressione: \[ \sqrt[3]{\frac{x^{2}}{y}} \cdot \sqrt[12]{\frac{27 x^{2}}{16 y}} \div \sqrt[6]{\frac{9 x^{5}}{4 y^{2}}} \] Espressiamo tutte le radici come potenze: \[ \left(\frac{x^{2}}{y}\right)^{\frac{1}{3}} \cdot \left(\frac{27 x^{2}}{16 y}\right)^{\frac{1}{12}} \div \left(\frac{9 x^{5}}{4 y^{2}}\right)^{\frac{1}{6}} \] Semplifichiamo ciascun termine separatamente: 1. **Primo termine:** \[ \left(\frac{x^{2}}{y}\right)^{\frac{1}{3}} = x^{\frac{2}{3}} \cdot y^{-\frac{1}{3}} \] 2. **Secondo termine:** \[ \left(\frac{27 x^{2}}{16 y}\right)^{\frac{1}{12}} = 27^{\frac{1}{12}} \cdot x^{\frac{2}{12}} \cdot 16^{-\frac{1}{12}} \cdot y^{-\frac{1}{12}} = 3^{\frac{3}{12}} \cdot x^{\frac{1}{6}} \cdot 2^{-\frac{4}{12}} \cdot y^{-\frac{1}{12}} = 3^{\frac{1}{4}} \cdot x^{\frac{1}{6}} \cdot 2^{-\frac{1}{3}} \cdot y^{-\frac{1}{12}} \] 3. **Terzo termine (denominatore):** \[ \left(\frac{9 x^{5}}{4 y^{2}}\right)^{\frac{1}{6}} = 9^{\frac{1}{6}} \cdot x^{\frac{5}{6}} \cdot 4^{-\frac{1}{6}} \cdot y^{-\frac{2}{6}} = 3^{\frac{2}{6}} \cdot x^{\frac{5}{6}} \cdot 2^{-\frac{2}{6}} \cdot y^{-\frac{1}{3}} = 3^{\frac{1}{3}} \cdot x^{\frac{5}{6}} \cdot 2^{-\frac{1}{3}} \cdot y^{-\frac{1}{3}} \] Ora, riscriviamo l'espressione completa combinando i termini: \[ \frac{ \left( x^{\frac{2}{3}} y^{-\frac{1}{3}} \right) \cdot \left( 3^{\frac{1}{4}} x^{\frac{1}{6}} 2^{-\frac{1}{3}} y^{-\frac{1}{12}} \right) }{ 3^{\frac{1}{3}} x^{\frac{5}{6}} 2^{-\frac{1}{3}} y^{-\frac{1}{3}} } \] Semplifichiamo gli esponenti: - **Per \( x \):** \[ \frac{2}{3} + \frac{1}{6} - \frac{5}{6} = \frac{4}{6} + \frac{1}{6} - \frac{5}{6} = 0 \] - **Per \( y \):** \[ - \frac{1}{3} - \frac{1}{12} + \frac{1}{3} = -\frac{1}{12} \] - **Per i coefficienti numerici:** \[ 3^{\frac{1}{4} - \frac{1}{3}} \cdot 2^{-\frac{1}{3} + \frac{1}{3}} = 3^{-\frac{1}{12}} \cdot 2^{0} = 3^{-\frac{1}{12}} \] Quindi, l'espressione si semplifica a: \[ \frac{1}{3^{\frac{1}{12}} \cdot y^{\frac{5}{12}}} \] **Risultato Finale:** \[ \frac{1}{3^{\frac{1}{12}} \, y^{\frac{5}{12}}} \] --- **Problema 7** Calcoliamo l'espressione: \[ \sqrt[8]{\frac{x^{3}+x^{2} y}{x-y}} \div \sqrt[4]{\frac{x^{2}+x y}{x-y}} \] Espressiamo le radici come potenze: \[ \left( \frac{x^{3} + x^{2} y}{x - y} \right)^{\frac{1}{8}} \div \left( \frac{x^{2} + x y}{x - y} \right)^{\frac{1}{4}} \] Semplifichiamo le frazioni nel numeratore e nel denominatore: 1. **Numeratore:** \[ x^{3} + x^{2} y = x^{2}(x + y) \] \[ \frac{x^{3} + x^{2} y}{x - y} = \frac{x^{2}(x + y)}{x - y} \] 2. **Denominatore:** \[ x^{2} + x y = x(x + y) \] \[ \frac{x^{2} + x y}{x - y} = \frac{x(x + y)}{x - y} \] Ora, riscriviamo l'espressione completa: \[ \left( \frac{x^{2}(x + y)}{x - y} \right)^{\frac{1}{8}} \div \left( \frac{x(x + y)}{x - y} \right)^{\frac{1}{4}} = \left( \frac{x^{2}(x + y)}{x - y} \right)^{\frac{1}{8}} \cdot \left( \frac{x(x + y)}{x - y} \right)^{-\frac{1}{4}} \] Espandiamo gli esponenti: \[ x^{\frac{2}{8}} \cdot (x + y)^{\frac{1}{8}} \cdot (x - y)^{-\frac{1}{8}} \cdot x^{-\frac{1}{4}} \cdot (x + y)^{-\frac{1}{4}} \cdot (x - y)^{\frac{1}{4}} \] Semplifichiamo gli esponenti: - **Per \( x \):** \[ \frac{2}{8} - \frac{1}{4} = \frac{1}{4} - \frac{1}{4} = 0 \] - **Per \( x + y \):** \[ \frac{1}{8} - \frac{1}{4} = \frac{1}{8} - \frac{2}{8} = -\frac{1}{8} \] - **Per \( x - y \):** \[ - \frac{1}{8} + \frac{1}{4} = -\frac{1}{8} + \frac{2}{8} = \frac{1}{8} \] Quindi, l'espressione si semplifica a: \[ \frac{(x - y)^{\frac{1}{8}}}{(x + y)^{\frac{1}{8}}} = \sqrt[8]{\frac{x - y}{x + y}} \] **Risultato Finale:** \[ \sqrt[8]{\frac{x - y}{x + y}} \]

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To solve these expressions, let's evaluate each step-by-step. **6. \( \sqrt[3]{\frac{x^{2}}{y}} \cdot \sqrt[12]{\frac{27 x^{2}}{16 y}}: \sqrt[6]{\frac{9 x^{5}}{4 y^{2}}} \)** First, rewrite the terms with exponents: 1. \( \sqrt[3]{\frac{x^{2}}{y}} = \left(\frac{x^{2}}{y}\right)^{\frac{1}{3}} = \frac{x^{\frac{2}{3}}}{y^{\frac{1}{3}}} \) 2. \( \sqrt[12]{\frac{27 x^{2}}{16 y}} = \left(\frac{27 x^{2}}{16 y}\right)^{\frac{1}{12}} = \frac{27^{\frac{1}{12}} x^{\frac{2}{12}}}{16^{\frac{1}{12}} y^{\frac{1}{12}}} \) Here, \( 27^{\frac{1}{12}} = 3^{\frac{3}{12}} = 3^{\frac{1}{4}} \) and \( 16^{\frac{1}{12}} = 2^{\frac{4}{12}} = 2^{\frac{1}{3}} \), so it simplifies to \( \frac{3^{\frac{1}{4}} x^{\frac{1}{6}}}{2^{\frac{1}{3}} y^{\frac{1}{12}}} \). 3. \( \sqrt[6]{\frac{9 x^{5}}{4 y^{2}}} = \left(\frac{9 x^{5}}{4 y^{2}}\right)^{\frac{1}{6}} = \frac{9^{\frac{1}{6}} x^{\frac{5}{6}}}{4^{\frac{1}{6}} y^{\frac{1}{3}}} \) Here, \( 9^{\frac{1}{6}} = 3^{\frac{2}{6}} = 3^{\frac{1}{3}} \) and \( 4^{\frac{1}{6}} = 2^{\frac{4}{6}} = 2^{\frac{2}{3}} \). Putting it all together, we have: \[ \text{Numerator: } \frac{x^{\frac{2}{3}}}{y^{\frac{1}{3}}} \cdot \frac{3^{\frac{1}{4}} x^{\frac{1}{6}}}{2^{\frac{1}{3}} y^{\frac{1}{12}}} \] The numerator becomes: \[ \frac{x^{\frac{2}{3} + \frac{1}{6}} \cdot 3^{\frac{1}{4}}}{y^{\frac{1}{3} + \frac{1}{12}} \cdot 2^{\frac{1}{3}}} \] To simplify exponents: 1. \( \frac{2}{3} = \frac{4}{6} \); thus \( \frac{2}{3} + \frac{1}{6} = \frac{5}{6} \) 2. For \( y: \frac{1}{3} + \frac{1}{12} = \frac{4}{12} + \frac{1}{12} = \frac{5}{12} \) So, the numerator is: \[ \frac{3^{\frac{1}{4}} x^{\frac{5}{6}}}{y^{\frac{5}{12}} \cdot 2^{\frac{1}{3}}} \] Now, dividing by the denominator gives: \[ \frac{\frac{3^{\frac{1}{4}} x^{\frac{5}{6}}}{y^{\frac{5}{12}} \cdot 2^{\frac{1}{3}}}}{\frac{3^{\frac{1}{3}} x^{\frac{5}{6}}}{2^{\frac{2}{3}} y^{\frac{2}{3}}}} \implies \frac{3^{\frac{1}{4} - \frac{1}{3}}}{2^{\frac{1}{3} - \frac{2}{3}}} \cdot \frac{y^{\frac{2}{3}}}{y^{\frac{5

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