Jerarquía con fracciones — Matemáticas 2

Video

Ejemplos

Ejemplo 1

$$\begin{aligned} \frac{1}{3} + \frac{4}{5} - \frac{6}{3} + \frac{3}{6} - \frac{6}{5} &= \\[14pt] -\frac{5}{3} - \frac{2}{5} + \frac{3}{6} &= \\[14pt] \frac{-50 - 12 + 15}{30} &= \mathbf{-\frac{47}{30}} \end{aligned}$$

Ejemplo 2

$$\begin{aligned} \left(\frac{3}{5}\right)^2 - 2\left(\frac{1}{3} - \frac{4}{5}\right) - \frac{3}{2} &= \\[14pt] \frac{9}{25} - \frac{2}{1}\left(\frac{5-12}{15}\right) - \frac{3}{2} &= \\[14pt] \frac{9}{25} - \frac{2}{1}\left(-\frac{7}{15}\right) - \frac{3}{2} &= \\[14pt] \frac{9}{25} + \frac{14}{15} - \frac{3}{2} &= \\[14pt] \frac{54 + 140 - 225}{150} &= \mathbf{-\frac{31}{150}} \end{aligned}$$

Ejemplo 3

$$\begin{aligned} \frac{1}{2}\left[\frac{3}{4} - 5\left(2 - \frac{1}{3}\right)\right] &= \\[14pt] \frac{1}{2}\left[\frac{3}{4} - \frac{5}{1}\left(\frac{5}{3}\right)\right] &= \\[14pt] \frac{1}{2}\left[\frac{3}{4} - \frac{25}{3}\right] &= \\[14pt] \frac{1}{2}\left[\frac{9 - 100}{12}\right] &= \mathbf{-\frac{91}{24}} \end{aligned}$$

Ejemplo 4

$$\begin{aligned} \frac{\dfrac{3}{2}\left[\dfrac{1}{4}-3\right]^2}{\dfrac{2}{5}-3\left(\dfrac{1}{4}\right)} &= \\[14pt] \frac{\dfrac{3}{2}\left[\dfrac{1-12}{4}\right]^2}{\dfrac{2}{5}-\dfrac{3}{4}} &= \\[14pt] \frac{\dfrac{3}{2}\left[\dfrac{-11}{4}\right]^2}{\dfrac{8-15}{20}} &= \\[14pt] \frac{\dfrac{3}{2}\left[\dfrac{121}{16}\right]}{-\dfrac{7}{20}} = \frac{\dfrac{363}{32}}{-\dfrac{7}{20}} &= \mathbf{-\frac{1815}{56}} \end{aligned}$$

Ejercicios

Bloque 1

$\dfrac{1}{4}-\dfrac{6}{5}+\dfrac{6}{4}+\dfrac{3}{5}-\dfrac{7}{10}=$

$\left(\dfrac{1}{2}\right)^3-4\left(\dfrac{3}{2}-\dfrac{5}{6}\right)-\dfrac{1}{4}=$

$\dfrac{2}{3}\left[\dfrac{1}{2}+4\left(3-\dfrac{2}{5}\right)\right]=$

$\dfrac{\dfrac{3}{5}\left[\dfrac{1}{2}-4\right]^2}{\dfrac{3}{2}-2\left(\dfrac{1}{6}\right)}=$

Bloque 2

$-\dfrac{2}{4}+\dfrac{6}{3}-\dfrac{8}{4}-\dfrac{7}{3}+\dfrac{1}{6}=$

$\left(\dfrac{2}{5}\right)^2-\dfrac{3}{5}-4\left(\dfrac{1}{2}-3\right)=$

$-\dfrac{2}{3}\left[\dfrac{4}{5}-6\left(2-\dfrac{1}{3}\right)\right]=$

$\dfrac{\dfrac{5}{3}\left[\dfrac{1}{3}-2\right]^2}{\dfrac{1}{5}-3\left(\dfrac{2}{6}\right)}=$

Bloque 3

$-\dfrac{8}{3}-\dfrac{2}{3}+\dfrac{7}{4}-\dfrac{9}{4}-\dfrac{3}{15}=$

$\dfrac{2}{3}-\left(\dfrac{1}{4}\right)^2-3\left(-\dfrac{1}{2}-\dfrac{3}{4}\right)=$

$-\dfrac{4}{5}\left[-\dfrac{2}{3}-\dfrac{2}{5}\left(\dfrac{3}{4}+1\right)\right]=$

$\dfrac{\dfrac{3}{4}\left[\dfrac{1}{5}-3\right]^2}{\dfrac{5}{2}-5\left(\dfrac{2}{4}\right)}=$

Bloque 4

$\dfrac{2}{6}-\dfrac{3}{5}+\dfrac{4}{2}-\dfrac{3}{5}-\dfrac{1}{2}=$

$\dfrac{3}{4}-\dfrac{2}{5}\left(\dfrac{1}{3}-\dfrac{4}{6}\right)+\left(\dfrac{2}{3}\right)^2=$

$\dfrac{6}{3}\left[4-5\left(-\dfrac{3}{2}-2\right)\right]=$

$\dfrac{-\dfrac{5}{6}\left[\dfrac{1}{2}-5\right]^2}{\dfrac{3}{4}-5\left(\dfrac{1}{10}\right)}=$

Respuestas

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Bloque 1

$$\begin{aligned} \tfrac{1}{4}-\tfrac{6}{5}+\tfrac{6}{4}+\tfrac{3}{5}-\tfrac{7}{10} &= \\[14pt] \tfrac{7}{4}-\tfrac{3}{5}-\tfrac{7}{10} &= \\[14pt] \tfrac{35-12-14}{20} &= \mathbf{\tfrac{9}{20}} \end{aligned}$$

$$\begin{aligned} \left(\tfrac{1}{2}\right)^3-4\left(\tfrac{3}{2}-\tfrac{5}{6}\right)-\tfrac{1}{4} &= \\[14pt] \tfrac{1}{8}-4\left(\tfrac{9-5}{6}\right)-\tfrac{1}{4} &= \\[14pt] \tfrac{1}{8}-\tfrac{16}{6}-\tfrac{1}{4} &= \\[14pt] \tfrac{3-64-6}{24} &= \mathbf{-\tfrac{67}{24}} \end{aligned}$$

$$\begin{aligned} \tfrac{2}{3}\!\left[\tfrac{1}{2}+4\!\left(3-\tfrac{2}{5}\right)\right] &= \\[14pt] \tfrac{2}{3}\!\left[\tfrac{1}{2}+4\!\left(\tfrac{13}{5}\right)\right] &= \\[14pt] \tfrac{2}{3}\!\left[\tfrac{1}{2}+\tfrac{52}{5}\right] &= \\[14pt] \tfrac{2}{3}\!\left[\tfrac{5+104}{10}\right] &= \mathbf{\tfrac{109}{15}} \end{aligned}$$

$$\begin{aligned} \frac{\tfrac{3}{5}\left[\tfrac{1}{2}-4\right]^2}{\tfrac{3}{2}-2\!\left(\tfrac{1}{6}\right)} &= \\[14pt] \frac{\tfrac{3}{5}\left[-\tfrac{7}{2}\right]^2}{\tfrac{3}{2}-\tfrac{2}{6}} &= \\[14pt] \frac{\tfrac{3}{5}\cdot\tfrac{49}{4}}{\tfrac{9-2}{6}} &= \\[14pt] \frac{\tfrac{147}{20}}{\tfrac{7}{6}} = \tfrac{882}{140} &= \mathbf{\tfrac{63}{10}} \end{aligned}$$

Bloque 2

$$\begin{aligned} -\tfrac{2}{4}+\tfrac{6}{3}-\tfrac{8}{4}-\tfrac{7}{3}+\tfrac{1}{6} &= \\[14pt] -\tfrac{10}{4}-\tfrac{1}{3}+\tfrac{1}{6} &= \\[14pt] \tfrac{-30-4+2}{12} &= \\[14pt] -\tfrac{32}{12} &= \mathbf{-\tfrac{8}{3}} \end{aligned}$$

$$\begin{aligned} \left(\tfrac{2}{5}\right)^2-\tfrac{3}{5}-4\!\left(\tfrac{1}{2}-3\right) &= \\[14pt] \tfrac{4}{25}-\tfrac{3}{5}-4\!\left(-\tfrac{5}{2}\right) &= \\[14pt] \tfrac{4}{25}-\tfrac{3}{5}+10 &= \\[14pt] \tfrac{8-30+500}{50} &= \mathbf{\tfrac{239}{25}} \end{aligned}$$

$$\begin{aligned} -\tfrac{2}{3}\!\left[\tfrac{4}{5}-6\!\left(2-\tfrac{1}{3}\right)\right] &= \\[14pt] -\tfrac{2}{3}\!\left[\tfrac{4}{5}-6\!\left(\tfrac{5}{3}\right)\right] &= \\[14pt] -\tfrac{2}{3}\!\left[\tfrac{4}{5}-10\right] &= \\[14pt] -\tfrac{2}{3}\!\left[-\tfrac{46}{5}\right] &= \mathbf{\tfrac{92}{15}} \end{aligned}$$

$$\begin{aligned} \frac{\tfrac{5}{3}\left[\tfrac{1}{3}-2\right]^2}{\tfrac{1}{5}-3\!\left(\tfrac{2}{6}\right)} &= \\[14pt] \frac{\tfrac{5}{3}\left[-\tfrac{5}{3}\right]^2}{\tfrac{1}{5}-1} &= \\[14pt] \frac{\tfrac{5}{3}\cdot\tfrac{25}{9}}{-\tfrac{4}{5}} &= \\[14pt] \frac{\tfrac{125}{27}}{-\tfrac{4}{5}} &= \mathbf{-\tfrac{625}{108}} \end{aligned}$$

Bloque 3

$$\begin{aligned} -\tfrac{8}{3}-\tfrac{2}{3}+\tfrac{7}{4}-\tfrac{9}{4}-\tfrac{3}{15} &= \\[14pt] -\tfrac{10}{3}-\tfrac{2}{4}-\tfrac{3}{15} &= \\[14pt] \tfrac{-200-30-12}{60} &= \mathbf{-\tfrac{121}{30}} \end{aligned}$$

$$\begin{aligned} \tfrac{2}{3}-\!\left(\tfrac{1}{4}\right)^2-3\!\left(-\tfrac{1}{2}-\tfrac{3}{4}\right) &= \\[14pt] \tfrac{2}{3}-\tfrac{1}{16}-3\!\left(-\tfrac{5}{4}\right) &= \\[14pt] \tfrac{2}{3}-\tfrac{1}{16}+\tfrac{15}{4} &= \\[14pt] \tfrac{32-3+180}{48} &= \mathbf{\tfrac{209}{48}} \end{aligned}$$

$$\begin{aligned} -\tfrac{4}{5}\!\left[-\tfrac{2}{3}-\tfrac{2}{5}\!\left(\tfrac{3}{4}+1\right)\right] &= \\[14pt] -\tfrac{4}{5}\!\left[-\tfrac{2}{3}-\tfrac{2}{5}\!\left(\tfrac{7}{4}\right)\right] &= \\[14pt] -\tfrac{4}{5}\!\left[-\tfrac{2}{3}-\tfrac{14}{20}\right] &= \\[14pt] -\tfrac{4}{5}\!\left[-\tfrac{82}{60}\right] &= \mathbf{\tfrac{82}{75}} \end{aligned}$$

$$\begin{aligned} \frac{\tfrac{3}{4}\left[\tfrac{1}{5}-3\right]^2}{\tfrac{5}{2}-5\!\left(\tfrac{2}{4}\right)} &= \\[14pt] \frac{\tfrac{3}{4}\left[-\tfrac{14}{5}\right]^2}{\tfrac{5}{2}-\tfrac{5}{2}} &= \\[14pt] \frac{\tfrac{3}{4}\cdot\tfrac{196}{25}}{0} &= \mathbf{\text{Indeterminado}} \end{aligned}$$

Bloque 4

$$\begin{aligned} \tfrac{2}{6}-\tfrac{3}{5}+\tfrac{4}{2}-\tfrac{3}{5}-\tfrac{1}{2} &= \\[14pt] \tfrac{1}{3}-\tfrac{6}{5}+\tfrac{3}{2} &= \\[14pt] \tfrac{10-36+45}{30} &= \mathbf{\tfrac{19}{30}} \end{aligned}$$

$$\begin{aligned} \tfrac{3}{4}-\tfrac{2}{5}\!\left(\tfrac{1}{3}-\tfrac{4}{6}\right)+\!\left(\tfrac{2}{3}\right)^2 &= \\[14pt] \tfrac{3}{4}-\tfrac{2}{5}\!\left(-\tfrac{1}{3}\right)+\tfrac{4}{9} &= \\[14pt] \tfrac{3}{4}+\tfrac{2}{15}+\tfrac{4}{9} &= \\[14pt] \tfrac{135+24+80}{180} &= \mathbf{\tfrac{239}{180}} \end{aligned}$$

$$\begin{aligned} \tfrac{6}{3}\!\left[4-5\!\left(-\tfrac{3}{2}-2\right)\right] &= \\[14pt] 2\!\left[4-5\!\left(-\tfrac{7}{2}\right)\right] &= \\[14pt] 2\!\left[4+\tfrac{35}{2}\right] &= \\[14pt] 2\!\left[\tfrac{43}{2}\right] &= \mathbf{43} \end{aligned}$$

$$\begin{aligned} \frac{-\tfrac{5}{6}\left[\tfrac{1}{2}-5\right]^2}{\tfrac{3}{4}-5\!\left(\tfrac{1}{10}\right)} &= \\[14pt] \frac{-\tfrac{5}{6}\left[-\tfrac{9}{2}\right]^2}{\tfrac{3}{4}-\tfrac{5}{10}} &= \\[14pt] \frac{-\tfrac{5}{6}\cdot\tfrac{81}{4}}{\tfrac{1}{4}} &= \\[14pt] \frac{-\tfrac{405}{24}}{\tfrac{1}{4}} = -\tfrac{1620}{24} &= \mathbf{-\tfrac{135}{2}} \end{aligned}$$

Ejercicios extra

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Bloque A

$\dfrac{7}{3} + \dfrac{3}{4} + \dfrac{6}{5} - \dfrac{8}{4} - \dfrac{2}{5} = \mathbf{\dfrac{127}{60}}$

$-\dfrac{3}{5} + \left(\dfrac{2}{5}\right)^2 - 4\!\left(-\dfrac{1}{3} - \dfrac{4}{5}\right) = \mathbf{\dfrac{307}{75}}$

$\dfrac{5}{6}\!\left[-\dfrac{3}{4} + \dfrac{3}{6}\!\left(\dfrac{4}{5} - 1\right)\right] = \mathbf{-\dfrac{17}{24}}$

$-\dfrac{4}{5} + \dfrac{3}{2} + \dfrac{4}{5} + \dfrac{3}{8} - \dfrac{6}{8} = \mathbf{\dfrac{9}{8}}$

$-\dfrac{6}{4} + \dfrac{4}{5} - \dfrac{5}{6} - \dfrac{7}{4} - \dfrac{1}{12} = \mathbf{-\dfrac{10}{3}}$

$\dfrac{4}{7} + \left(-\dfrac{3}{6}\right)^2 - 5\!\left(-\dfrac{2}{4} + \dfrac{5}{6}\right) = \mathbf{-\dfrac{23}{28}}$

$-\dfrac{6}{7}\!\left[-\dfrac{4}{5} - \dfrac{4}{7}\!\left(\dfrac{5}{6} + 1\right)\right] = \mathbf{\dfrac{6}{5}}$

$\dfrac{5}{8} - \left(\dfrac{4}{7}\right)^2 + 6\!\left(-\dfrac{3}{5} + \dfrac{6}{7}\right) = \mathbf{\dfrac{717}{392}}$

Bloque B

$\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{8} = \mathbf{\dfrac{5}{8}}$

$\dfrac{3}{5} \div \dfrac{2}{3} + \dfrac{1}{10} = \mathbf{1}$

$\left(\dfrac{2}{3} - \dfrac{1}{4}\right) \cdot \dfrac{12}{5} = \mathbf{1}$

$\dfrac{4}{3} - \left[\dfrac{1}{2}\cdot\left(\dfrac{1}{4} + \dfrac{1}{3}\right)\right] = \mathbf{\dfrac{25}{24}}$

$\dfrac{\dfrac{3}{4} - \dfrac{1}{2}}{\dfrac{1}{8} + \dfrac{1}{4}} = \mathbf{\dfrac{2}{3}}$

$\left(\dfrac{1}{2}\right)^2 + \dfrac{3}{4}\cdot\dfrac{1}{3} = \mathbf{\dfrac{1}{2}}$

$\dfrac{2}{5}\cdot\left(\dfrac{1}{2} + \dfrac{1}{3}\right) - \dfrac{1}{10} = \mathbf{\dfrac{7}{30}}$

$\dfrac{5}{6} \div \left(1 - \dfrac{1}{3}\right) + \dfrac{1}{4} = \mathbf{\dfrac{3}{2}}$