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(3) RAZONAMIENTO. Cdícular las siguiențes raices aplicando las propiedades de los'radicales: 13. \( \sqrt{(-5)^{2}} \) 14. \( \sqrt[3]{-729 x^{6}} \) 15. \( \sqrt[4]{16 a^{8} b^{4}} \) 16. \( \sqrt[5]{3.125 m^{10} n^{5}} \) 17. \( \sqrt[2]{64 x^{10} y^{8} z^{6}} \) 18. \( \sqrt[3]{-27 x^{-6} y^{9}} \) 19. \( \sqrt{\frac{9 y^{8}}{m^{12}}} \) 20. \( \frac{\sqrt{12 a^{4} b^{6}}}{1 \sqrt{3 b^{2}}} \) \( : 2 \pi \cdot \sqrt[5]{\frac{128 x^{15} y^{20}}{4 z^{10}}} \) 22. \( \sqrt[3]{4 \sqrt{4 m^{6}}} \) 23. \( \sqrt[4]{\sqrt[2]{m^{16} y^{24} z^{8}}} \) 24. \( \sqrt{x^{2} \sqrt[3]{64 y^{6}}} \) - MODELACIÓN: Escribir, las siguientes raíces en forma de potencia y simplificar cada resultado si es posible. 25. \( \sqrt[5]{x^{5} y^{10}} \) 26. \( \sqrt[3]{w^{3} z^{3}} \) 27. \( \sqrt{4 y^{4}} \) 28. \( \sqrt[6]{64 x^{12} y^{6}} \) 29. \( \sqrt[n]{m^{2 n^{n}}} \) 30. \( \sqrt[4]{7^{8} x^{16} y^{8}} \) 31. \( \sqrt[10]{10^{10} \chi_{i}^{20}} \) 32. \( \sqrt[3]{27 x^{3}} \) 33. \( \sqrt[5]{243 x^{10}} \)

Ask by Hardy Love. in Colombia
Mar 10,2025

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**13.** \[ \sqrt{(-5)^2} = 5 \] **14.** \[ \sqrt[3]{-729\,x^6} = -9x^2 \] **15.** \[ \sqrt[4]{16\,a^8\,b^4} = 2a^2b \] **16.** \[ \sqrt[5]{3.125\,m^{10}\,n^5} = \frac{\sqrt[5]{25}\,m^2\,n}{2^{3/5}} \] **17.** \[ \sqrt{64\,x^{10}\,y^8\,z^6} = 8x^5y^4z^3 \]

Solution

**13.** Calcular \[ \sqrt{(-5)^2}. \] Utilizamos que \(\sqrt{a^2}=|a|\). Entonces: \[ \sqrt{(-5)^2}=|{-5}|=5. \] **14.** Calcular \[ \sqrt[3]{-729 x^{6}}. \] Notamos que \[ -729=(-9)^3, \] y además \[ x^6=(x^2)^3. \] Por lo que: \[ \sqrt[3]{-729\,x^6}=\sqrt[3]{(-9)^3\, (x^2)^3}=-9x^2. \] **15.** Calcular \[ \sqrt[4]{16\,a^{8}\,b^{4}}. \] Observamos que \(16=2^4\) y escribimos el radicando como: \[ 2^4\,a^8\,b^4. \] Aplicando la raíz: \[ \sqrt[4]{2^4\,a^8\,b^4}=2\,a^{8/4}\,b^{4/4}=2a^2b. \] **16.** Calcular \[ \sqrt[5]{3.125\,m^{10}\,n^{5}}. \] Notamos que \(3.125=\frac{25}{8}\) (ya que \(\frac{25}{8}=3.125\)) y escribimos: \[ \sqrt[5]{\frac{25}{8}\,m^{10}\,n^5}= \frac{\sqrt[5]{25\,m^{10}\,n^5}}{\sqrt[5]{8}}. \] Observamos que: \[ m^{10}=(m^2)^5 \quad \text{y} \quad n^5=(n)^5. \] Entonces, \[ \sqrt[5]{25\,m^{10}\,n^5}=\sqrt[5]{25}\,m^2\,n \] y \[ \sqrt[5]{8}=8^{1/5}=2^{3/5}. \] La expresión queda: \[ \frac{\sqrt[5]{25}\,m^2\,n}{2^{3/5}}. \] **17.** Calcular \[ \sqrt{64\,x^{10}\,y^{8}\,z^{6}}. \] Aplicamos la raíz cuadrada término a término: - \(\sqrt{64}=8\), - \(\sqrt{x^{10}}=x

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¡Vamos a desmenuzar esas raíces y hacer las cosas más claras y entretenidas! **13.** \( \sqrt{(-5)^{2}} = \sqrt{25} = 5 \) **14.** \( \sqrt[3]{-729 x^{6}} = -9x^{2} \) **15.** \( \sqrt[4]{16 a^{8} b^{4}} = 2 a^{2} b \) **16.** \( \sqrt[5]{3.125 m^{10} n^{5}} = 5 \cdot m^{2} \cdot n \) **17.** \( \sqrt[2]{64 x^{10} y^{8} z^{6}} = 8 x^{5} y^{4} z^{3} \) **18.** \( \sqrt[3]{-27 x^{-6} y^{9}} = -3 y^{3} x^{-2} \) **19.** \( \sqrt{\frac{9 y^{8}}{m^{12}}} = \frac{3 y^{4}}{m^{6}} \) **20.** \( \frac{\sqrt{12 a^{4} b^{6}}}{1 \sqrt{3 b^{2}}} = \frac{2 a^{2} b^{3}}{\sqrt{3}} \) **21.** \( 2 \pi \cdot \sqrt[5]{\frac{128 x^{15} y^{20}}{4 z^{10}}} = 2 \pi \cdot \sqrt[5]{32 x^{15} y^{20} z^{-10}} \) **22.** \( \sqrt[3]{4 \sqrt{4 m^{6}}} = \sqrt[3]{4 \cdot 2 m^{3}} = \sqrt[3]{8 m^{3}} = 2 m \) **23.** \( \sqrt[4]{\sqrt[2]{m^{16} y^{24} z^{8}}} = \sqrt[4]{m^{8} y^{12} z^{4}} = m^{2} y^{3} z \) **24.** \( \sqrt{x^{2} \sqrt[3]{64 y^{6}}} = \sqrt{x^{2} 4y^{2}} = 2xy \) ¡Ahora, vamos a modelar esas raíces en forma de potencia y simplificar! **25.** \( \sqrt[5]{x^{5} y^{10}} = x^{1} y^{2} \) **26.** \( \sqrt[3]{w^{3} z^{3}} = w^{1} z^{1} \) **27.** \( \sqrt{4 y^{4}} = 2 y^{2} \) **28.** \( \sqrt[6]{64 x^{12} y^{6}} = 2 x^{2} y^{1} \) **29.** \( \sqrt[n]{m^{2 n^{n}}} = m^{\frac{2}{n}} \) **30.** \( \sqrt[4]{7^{8} x^{16} y^{8}} = 7^{2} x^{4} y^{2} \) **31.** \( \sqrt[10]{10^{10} \chi_{i}^{20}} = 10^{1} \chi_{i}^{2} \) **32.** \( \sqrt[3]{27 x^{3}} = 3 x \) **33.** \( \sqrt[5]{243 x^{10}} = 3 \cdot x^{2} \) ¡Y ahí lo tienes! Cada raíz desenterrada y presentada en su forma más pura. Si tienes más preguntas, estoy aquí para ayudar.

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