a) \dfrac{2}{45}-\dfrac{11}{21}+\dfrac{5}{14}
b) \dfrac{4}{7} \cdot \dfrac{21}{9}
c) \dfrac{3}{2} \div \dfrac{6}{18}
d) \left(-\dfrac{5}{4}\right)^{2} \div \dfrac{15}{2}-\dfrac{1}{2}
SOLUCIÓN.
a)
\dfrac{2}{45}-\dfrac{11}{21}+\dfrac{5}{14} \quad \overset{\text{m.c.m}(45,21,14)=630}{=} \quad \dfrac{2 \cdot 630 \div 45}{630}+\dfrac{(-11)\cdot 630 \div 21}{630}+\dfrac{5 \cdot 630\div 14}{630}
=\dfrac{28}{630}+\dfrac{(-330)}{630+\dfrac{225}{630}}
=\dfrac{28+(-330)+225}{630}
=\dfrac{-77}{630}
=-\dfrac{77}{630}
\quad \overset{\text{m.c.d.}(77,630)=7}{=}\quad -\dfrac{11}{90}
b)
\dfrac{4}{7} \cdot \dfrac{21}{9}
=\dfrac{4 \cdot 21}{7 \cdot 9}
=\dfrac{21 \cdot 4}{7 \cdot 9}
=\dfrac{21}{7}\cdot \dfrac{4}{9}
=3\cdot \dfrac{4}{9}
=\dfrac{3\cdot 4}{9}
=\dfrac{4\cdot 3}{9}
=4 \cdot \dfrac{3}{9}
=4 \cdot \dfrac{1}{3}
=\dfrac{4}{3}
c)
\dfrac{3}{2} \div \dfrac{6}{18}
\overset{\frac{6}{18}=\frac{1}{3}}{=} \quad \dfrac{3}{2} \div \dfrac{1}{3}
=\dfrac{3}{2} \cdot \text{inverso} \left( \dfrac{1}{3} \right)
=\dfrac{3}{2} \cdot 3
=\dfrac{3 \cdot 3}{2}
=\dfrac{9}{2}
d)
\left(-\dfrac{5}{4}\right)^{2} \div \dfrac{15}{2}-\dfrac{1}{2}
=\dfrac{25}{16} \div \dfrac{15}{2}-\dfrac{1}{2}
=\dfrac{25}{16} \cdot \text{inverso}\left( \dfrac{15}{2}\right)-\dfrac{1}{2}
=\dfrac{25}{16} \cdot \dfrac{2}{15}-\dfrac{1}{2}
=\dfrac{25 \cdot 2}{16 \cdot 15} -\dfrac{1}{2}
=\dfrac{2 \cdot 25}{16 \cdot 15} -\dfrac{1}{2}
=\dfrac{2}{16}\cdot \dfrac{25}{15} -\dfrac{1}{2}
=\dfrac{1}{8}\cdot \dfrac{5}{3} -\dfrac{1}{2}
=\dfrac{1\cdot 5}{8 \cdot 3} -\dfrac{1}{2}
=\dfrac{5}{24} -\dfrac{1}{2}
=\dfrac{5}{24} -\dfrac{12}{24}
=\dfrac{5}{24} +\dfrac{(-12)}{24}
=\dfrac{5+(-12)}{24}
=\dfrac{(-7)}{24}
=-\dfrac{7}{24}
\square
[autoría]
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