FRACCIONES...

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Flashcards by , created about 5 years ago

Suma y Resta de FRACCIONES: a/b ± c/d = (ad ± bc) / bd Con reducción, e impropia (>1) en mixto.

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JL Cadenas
Created by JL Cadenas about 5 years ago
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Question Answer
\(\frac {2}{3}\) + \(\frac {3}{4}\) = = \(\frac {8+9}{3·4}\)=\(\frac {17}{12}\)=\(1\frac {5}{12}\)
\(\frac {1}{4}\) + \(\frac {2}{3}\) = = \(\frac {3+8}{4·3}\)=\(\frac {11}{12}\)
\(\frac {1}{2}\) + \(\frac {3}{4}\) = = \(\frac {4+6}{2·4}\)=\(\frac {5}{4}\)=\(1\frac {1}{4}\)
\(\frac {1}{2}\) + \(\frac {1}{3}\) = = \(\frac {3+2}{2·3}\)=\(\frac {5}{6}\)
\(\frac {2}{3}\) + \(\frac {1}{2}\) = = \(\frac {4+3}{3·2}\)=\(\frac {7}{6}\)=\(1\frac {1}{6}\)
\(\frac {1}{3}\) + \(\frac {3}{4}\) = = \(\frac {4+9}{3·4}\)=\(\frac {13}{12}\)=\(1\frac {1}{12}\)
\(\frac {3}{4}\) + \(\frac {1}{5}\) = = \(\frac {15+4}{4·5}\)=\(\frac {19}{20}\)
\(\frac {2}{5}\) + \(\frac {1}{3}\) = = \(\frac {6+5}{5·3}\)=\(\frac {11}{15}\)
\(\frac {2}{5}\) + \(\frac {1}{2}\) = = \(\frac {4+5}{5·2}\)=\(\frac {9}{10}\)
\(\frac {1}{3}\) + \(\frac {1}{5}\) = = \(\frac {5+3}{3·5}\)=\(\frac {8}{15}\)
\(\frac {2}{3}\) + \(\frac {3}{5}\) = = \(\frac {10+9}{3·5}\)=\(\frac {19}{15}\)=\(1\frac {4}{15}\)
\(\frac {3}{5}\) + \(\frac {1}{3}\) = = \(\frac {9+5}{5·3}\)=\(\frac {14}{15}\)
\(\frac {3}{5}\) – \(\frac {1}{3}\) = = \(\frac {9–5}{5·3}\)=\(\frac {4}{15}\)
\(\frac {2}{3}\) – \(\frac {1}{4}\) = = \(\frac {8–3}{3·4}\)=\(\frac {5}{12}\)
\(\frac {3}{4}\) – \(\frac {2}{3}\) = = \(\frac {9–8}{4·3}\)=\(\frac {1}{12}\)
\(\frac {3}{4}\) – \(\frac {1}{2}\) = = \(\frac {6–4}{4·2}\)=\(\frac {2}{8}\)=\(\frac {1}{4}\)
\(\frac {1}{2}\) – \(\frac {1}{3}\) = = \(\frac {3–2}{2·3}\)=\(\frac {1}{6}\)
\(\frac {2}{3}\) – \(\frac {1}{2}\) = = \(\frac {4–3}{3·2}\)=\(\frac {1}{6}\)
\(\frac {3}{4}\) – \(\frac {1}{3}\) = = \(\frac {9–4}{4·3}\)=\(\frac {5}{12}\)
\(\frac {3}{4}\) – \(\frac {1}{5}\) = = \(\frac {15–4}{4·5}\)=\(\frac {11}{20}\)
\(\frac {2}{5}\) – \(\frac {1}{3}\) = = \(\frac {6–5}{5·3}\)=\(\frac {1}{15}\)
\(\frac {1}{2}\) – \(\frac {2}{5}\) = = \(\frac {5–4}{2·5}\)=\(\frac {1}{10}\)
\(\frac {1}{3}\) – \(\frac {1}{5}\) = = \(\frac {5–3}{3·5}\)=\(\frac {2}{15}\)
\(\frac {2}{3}\) – \(\frac {3}{5}\) = = \(\frac {10–9}{3·5}\)=\(\frac {1}{15}\)