- Ejercicio 1 (2 puntos)
Resolver las siguientes operaciones con fracciones, simplificando en todo momento los pasos intermedios y el resultado.
\(\displaystyle\frac{\displaystyle\left(\frac{2}{5}-\frac{1}{3}\right):\left(1-\frac{1}{3}\cdot\frac{6}{5}\right)}{\displaystyle1-\left(\frac{2}{3}\cdot\frac{1}{3}-1\right)}+1\)
\(\displaystyle\frac{\displaystyle3+\frac{3}{2+1/2}}{\displaystyle\frac{1}{15}+\left(\frac{3}{2}-\frac{1}{3}\cdot\frac{5}{2}\right)\cdot\frac{1}{5}}\)
La solución aquí
La solución aquí
\(\displaystyle\frac{\displaystyle\left(\frac{2}{5}-\frac{1}{3}\right):\left(1-\frac{1}{3}\cdot\frac{6}{5}\right)}{\displaystyle1-\left(\frac{2}{3}\cdot\frac{1}{3}-1\right)}+1= \frac{\displaystyle\left(\frac{6}{15}-\frac{5}{15}\right):\left(1-\frac{6}{15}\right)}{\displaystyle1-\left(\frac{2}{9}-1\right)}+1=\)
\(\displaystyle=\frac{\displaystyle\left(\frac{1}{15}\right):\left(\frac{9}{15}\right)}{\displaystyle1-\left(-\frac{7}{9}\right)}+1=\frac{\displaystyle\frac{15}{15\cdot9}}{\displaystyle1+\frac{7}{9}}+1=\frac{\displaystyle\frac{1}{9}}{\displaystyle\frac{16}{9}}+1=\frac{9}{9\cdot16}+1=\)
\(\displaystyle=\frac{1}{16}+1=\frac{17}{16}\)
\(\displaystyle\frac{\displaystyle3+\frac{3}{2+1/2}}{\displaystyle\frac{1}{15}+\left(\frac{3}{2}-\frac{1}{3}\cdot\frac{5}{2}\right)\cdot\frac{1}{5}}= \frac{\displaystyle3+\frac{3}{5/2}}{\displaystyle\frac{1}{15}+\left(\frac{3}{2}-\frac{5}{6}\right)\cdot\frac{1}{5}}=\)
\(=\frac{\displaystyle3+\frac{6}{5}}{\displaystyle\frac{1}{15}+\left(\frac{9}{6}-\frac{5}{6}\right)\cdot\frac{1}{5}}=\displaystyle\frac{\displaystyle\frac{15}{5}+\frac{6}{5}}{\displaystyle\frac{1}{15}+\frac{4}{6}\cdot\frac{1}{5}}= \frac{\displaystyle\frac{21}{5}}{\displaystyle\frac{1}{15}+\frac{4}{30}}=\frac{\displaystyle\frac{21}{5}}{\displaystyle\frac{2}{30}+\frac{4}{30}}=\)
\(\displaystyle=\frac{\displaystyle\frac{21}{5}}{\displaystyle\frac{6}{30}}=\frac{21\cdot30}{5\cdot6}=\frac{21\cdot30}{30}=21\)
- Ejercicio 2 (2 puntos)
Realiza las siguientes operaciones con potencias y simplifica el resultado todo lo posible (se puede dejar el resultado en forma de potencia).
\(\displaystyle\frac{4^2\cdot2^{-2}\cdot9^{-3}\cdot6^{3}}{12\cdot3^{-3}\cdot2\cdot3^{-3}}\)
\(\displaystyle\frac{\displaystyle\left(\frac{5}{4}\right)^{-3}\cdot\left(\frac{25}{4}\right)^3}{\displaystyle5^3\cdot\left(\frac{2}{5}\right)^{-2}\cdot\left(\frac{5}{2}\right)^{-3}\cdot\left(\frac{4}{5}\right)^2}\)
La solución aquí
La solución aquí
\(\displaystyle\frac{4^2\cdot2^{-2}\cdot9^{-3}\cdot6^{3}}{12\cdot3^{-3}\cdot2\cdot3^{-3}}=\frac{(2^2)^2\cdot2^{-2}\cdot(3^2)^{-3}\cdot(2\cdot3)^{3}}{(2^2\cdot3)\cdot3^{-3}\cdot2\cdot3^{-3}}=\)
\(\displaystyle=\frac{2^4\cdot2^{-2}\cdot3^{-6}\cdot2^{3}\cdot3^3}{2^2\cdot3\cdot3^{-3}\cdot2\cdot3^{-3}}=\frac{2^5\cdot3^{-3}}{2^3\cdot3^{-5}}=2^2\cdot3^2=4\cdot9=36\)
\(\displaystyle\frac{\displaystyle\left(\frac{5}{4}\right)^{-3}\cdot\left(\frac{25}{4}\right)^3}{\displaystyle5^3\cdot\left(\frac{2}{5}\right)^{-2}\cdot\left(\frac{5}{2}\right)^{-3}\cdot\left(\frac{4}{5}\right)^2}=\frac{\displaystyle\left(\frac{4}{5}\right)^{3}\cdot\left(\frac{25}{4}\right)^3}{\displaystyle5^3\cdot\left(\frac{5}{2}\right)^{2}\cdot\left(\frac{2}{5}\right)^{3}\cdot\left(\frac{4}{5}\right)^2}=\)
\(\displaystyle=\frac{\displaystyle\frac{2^6}{5^3}\cdot\frac{5^6}{2^6}}{\displaystyle5^3\cdot\frac{5^2}{2^2}\cdot\frac{2^3}{5^3}\cdot\frac{2^4}{5^2}}=\frac{5^3}{2^5}=\frac{125}{32}\)
- Ejercicio 3 (1 punto)
Opera y simplifica extrayendo factores siempre que sea posible (recuerda que has de factorizar los números que no sean primos).
\(\displaystyle\sqrt{16\sqrt[5]{64}}\)
\(\displaystyle3\sqrt{2}+4\sqrt{8}-\sqrt{32}+\sqrt{50}\)
La solución aquí
La solución aquí
\(\displaystyle\sqrt{16\sqrt[5]{64}}=\sqrt{2^4\sqrt[5]{2^6}}=\sqrt{\sqrt[5]{(2^4)^5\cdot2^6}}=\sqrt[10]{2^{20}\cdot2^6}=\sqrt[10]{2^{26}}=\)
\(\displaystyle=2^2\,\sqrt[10]{2^6}=4\,\sqrt[5]{2^3}=4\,\sqrt[5]{8}\)
\(\displaystyle3\sqrt{2}+4\sqrt{8}-\sqrt{32}+\sqrt{50}=3\sqrt{2}+4\sqrt{2^3}-\sqrt{2^5}+\sqrt{2\cdot5^2}=\)
\(\displaystyle=3\sqrt{2}+4\cdot2\,\sqrt{2}-2^2\,\sqrt{2}+5\,\sqrt{2}=(3+8-4+5)\sqrt{2}=12\,\sqrt{2}\)
- Ejercicio 4 (1 punto)
Racionalizar las siguientes expresiones y simplificar el resultado en la medida de lo posible.
\(\displaystyle \frac{6}{\sqrt[3]{3}}\)
\(\displaystyle \frac{9}{\sqrt{5}-\sqrt{2}}\)
La solución aquí
La solución aquí
\(\displaystyle \frac{6}{\sqrt[3]{3}}=\frac{6\cdot\sqrt[3]{3}}{\sqrt[3]{3}\cdot\sqrt[3]{3^2}}=\frac{6\cdot\sqrt[3]{3^2}}{\sqrt[3]{3^3}}=\frac{6\cdot\sqrt[3]{9}}{3}=2\sqrt[3]{9}\)
\(\displaystyle \frac{9}{\sqrt{5}-\sqrt{2}}=\frac{9\cdot\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\cdot\left(\sqrt{5}+\sqrt{2}\right)}=\)
\(\displaystyle =\frac{9\cdot\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}^2-\sqrt{2}^2}=\frac{9\cdot\left(\sqrt{5}+\sqrt{2}\right)}{5-2}=\)
\(\displaystyle =\frac{9\cdot\left(\sqrt{5}+\sqrt{2}\right)}{3}=3\cdot\left(\sqrt{5}+\sqrt{2}\right)\)
- Ejercicio 5 (4 puntos)
Resolver las siguientes ecuaciones.
\(\displaystyle \frac{1}{3}(x+2)-\frac{1}{5}(2x-3)=4-\frac{2x}{15}\)
\(\displaystyle \frac{x+5}{3}+\frac{x-3}{2}=\frac{x+5}{5}-\frac{3x}{15}\)
\(\displaystyle \frac{x(x+1)}{5}=2x^2-4x\)
\(\displaystyle \left(\frac{3}{2}x-2\right)^2-(x-1)(x+1)=-2\)
La solución aquí
La solución aquí
\(\displaystyle \frac{1}{3}(x+2)-\frac{1}{5}(2x-3)=4-\frac{2x}{15}\Rightarrow5(x+2)-3(2x-3)=60-2x\Rightarrow\)
\(\Rightarrow 5x+10-6x+9=60-2x\Rightarrow5x-6x+2x=60-10-9\Rightarrow\)
\(\Rightarrow x=41\)
\(\displaystyle \frac{x+5}{3}+\frac{x-3}{2}=\frac{x+5}{5}-\frac{3x}{15}\Rightarrow\)
\(\Rightarrow10(x+5)+15(x-3)=6(x+5)-6x\Rightarrow\)
\(\Rightarrow10x+50+15x-45=6x+30-6x\Rightarrow\)
\(\Rightarrow10x+15x-6x+6x=30-50+45\Rightarrow\)
\(\displaystyle\Rightarrow25x=25\Rightarrow x=\frac{25}{25}\Rightarrow x=1\)
\(\displaystyle \frac{x(x+1)}{5}=2x^2-4x\)
\(\displaystyle \left(\frac{3}{2}x-2\right)^2-(x-1)(x+1)=-2\)