a) \dfrac{2}{3}+\dfrac{5}{18}-\dfrac{1}{12}
b) \dfrac{4}{5}\cdot \dfrac{25}{8}
c) \dfrac{3}{4}\div \dfrac{9}{2}
d) \left(1-\dfrac{5}{2}\right)^3
SOLUCIÓN.
a)
\dfrac{2}{3}+\dfrac{5}{18}-\dfrac{1}{12}=
\overset{\text{m.c.m}(3,18,12)=36}{=} \quad \quad \dfrac{2 \cdot 36 \div 3}{36}+\dfrac{5 \cdot 36 \div 18}{36}+\dfrac{(-1) \cdot 36 \div 12 }{36}
=\dfrac{24}{36}+\dfrac{10}{36}+\dfrac{(-3)}{36}
=\dfrac{24+10+(-3)}{36}
=\dfrac{31}{36}
b)
\dfrac{4}{5}\cdot \dfrac{25}{8}
=\dfrac{4 \cdot 25}{5 \cdot 8}
=\dfrac{25 \cdot 4}{5 \cdot 8}
=\dfrac{25}{5} \cdot \dfrac{4}{8}
=5 \cdot \dfrac{1}{2}
=\dfrac{5}{2}
c)
\dfrac{3}{4}\div \dfrac{9}{2}
=\dfrac{3}{4}\cdot \text{inverso}\left( \dfrac{9}{2} \right)
=\dfrac{3}{4}\cdot \dfrac{2}{9}
=\dfrac{3 \cdot 2}{4 \cdot 9}
=\dfrac{2 \cdot 3}{4 \cdot 9}
=\dfrac{2}{4}\cdot \dfrac{3}{9}
=\dfrac{1}{2}\cdot \dfrac{1}{3}
=\dfrac{1 \cdot 1 }{2 \cdot 3}
=\dfrac{1 }{6}
d)
\left(1-\dfrac{5}{2}\right)^3
=\left(\dfrac{2}{2}-\dfrac{5}{2}\right)^3
=\left(\dfrac{2}{2}+\dfrac{(-5)}{2}\right)^3
=\left(\dfrac{2+(-5)}{2}\right)^3
=\left(\dfrac{(-3)}{2}\right)^3
=\dfrac{(-3)^3}{2^3}
=\dfrac{(-27)}{8}
=-\dfrac{27}{8}
\square
[autoría]
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