Vom scrie sub forma unui singur exercițiu:
[tex]a) \ \bigg(\dfrac{^{5)} 2}{3} + \dfrac{^{3)}1}{5}\bigg) \cdot \bigg(\dfrac{^{5)} 2}{3} - \dfrac{^{3)}1}{5}\bigg) = \dfrac{10+3}{15} \cdot \dfrac{10-3}{15} = \dfrac{13}{15} \cdot \dfrac{7}{15} = \dfrac{91}{225}[/tex]
Sau putem utiliza formula de calcul prescurtat:
[tex]\bigg(\dfrac{2}{3} + \dfrac{1}{5}\bigg) \cdot \bigg(\dfrac{2}{3} - \dfrac{1}{5}\bigg) = \bigg(\dfrac{2}{3}\bigg)^2 - \bigg(\dfrac{1}{5}\bigg)^2 = \dfrac{^{25)} 4}{9} - \dfrac{^{9)} 1}{25} = \dfrac{100 - 9}{225} = \dfrac{91}{225}[/tex]
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[tex]b) \ \bigg(\dfrac{5}{2} + \dfrac{11}{12}\bigg) \cdot \bigg(\dfrac{5}{2} - \dfrac{11}{12}\bigg) = \bigg(\dfrac{5}{2}\bigg)^2 - \bigg(\dfrac{11}{12}\bigg)^2 = \dfrac{^{36)} 25}{4} - \dfrac{121}{144} = \dfrac{900 - 121}{1445} = \dfrac{779}{144}[/tex]
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[tex]c) \ \bigg(\dfrac{5}{8} + \dfrac{^{2)}1}{4}\bigg) \cdot \bigg(\dfrac{5}{8} - \dfrac{^{2)}1}{4}\bigg) = \dfrac{5 + 2}{8} \cdot \dfrac{5 - 2}{8} = \dfrac{7}{8} \cdot \dfrac{3}{8} = \dfrac{21}{64}[/tex]
