SOLUCIÓN.
\dfrac{5}{9}-\left(-\dfrac{3}{4}+\dfrac{1}{2}\right)+\dfrac{10}{3}\cdot \left(\dfrac{1}{2}-\dfrac{1}{5}\right)^2=
=\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{10}{3}\cdot \left(\dfrac{3}{10} \right)^2
=\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{10}{3}\cdot \dfrac{9}{100}
=\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{3}{10}
=\dfrac{5\cdot 20}{180}-\dfrac{1 \cdot 45}{180}+\dfrac{3 \cdot 18}{180} ( reduciendo a común denominador, \text{m.c.m}(9,4,10)=180 )
=\dfrac{100}{180}-\dfrac{45}{180}+\dfrac{54}{180}
=\dfrac{100-45+54}{180}
=\dfrac{109}{180}
\square
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