Cálculos combinados | Práctica 0 ejercicio 1a

EJERCICIO 1:Calcule
(a) $\dfrac{2}{3}-\left(-\dfrac{1}{2}+1-(-\dfrac{1}{2}-\dfrac{5}{12})-2\right) -\dfrac{1}{4}-
\left(-1-(2\cdot \dfrac{1}{6}-\dfrac{1}{4})+\dfrac{2}{3}\right)$


Solución:
(a) $\dfrac{2}{3}-\left(-\dfrac{1}{2}+1-(-\dfrac{1}{2}-\dfrac{5}{12})-2\right) -\dfrac{1}{4}-
\left(-1-(2\cdot \dfrac{1}{6}-\dfrac{1}{4})+\dfrac{2}{3}\right)=$

$\dfrac{2}{3}-\left(-\dfrac{1}{2}+1-(\dfrac{-6-5}{12})-2\right) -\dfrac{1}{4}-
\left(-1-(\cancel{2}\cdot \dfrac{1}{\cancel{6}_3}-\dfrac{1}{4})+\dfrac{2}{3}\right)=$

$\dfrac{2}{3}-\left(-\dfrac{1}{2}+1-(\dfrac{-11}{12})-2\right) -\dfrac{1}{4}-
\left(-1-( \dfrac{1}{3}-\dfrac{1}{4})+\dfrac{2}{3}\right)=$

$\dfrac{2}{3}-\left(-\dfrac{1}{2}+1-(\dfrac{-11}{12})-2\right) -\dfrac{1}{4}-
\left(-1-( \dfrac{4-3}{12})+\dfrac{2}{3}\right)=$

$\dfrac{2}{3}-\left(-\dfrac{1}{2}+1+\dfrac{11}{12}-2\right) -\dfrac{1}{4}-
\left(-1-\dfrac{1}{12}+\dfrac{2}{3}\right)=$

$\dfrac{2}{3}-\left(\dfrac{-6+12+11-24}{12}\right) -\dfrac{1}{4}-
\left(\dfrac{-12-1+8}{12}\right)=$

$\dfrac{2}{3}-\left(\dfrac{-7}{12}\right) -\dfrac{1}{4}-
\left(\dfrac{-5}{12}\right)=$

$\dfrac{2}{3}+\dfrac{7}{12} -\dfrac{1}{4}+\dfrac{-5}{12}=$

$\dfrac{8+7-3+5}{12}= \boxed{\dfrac{17}{12}}$