Question : Calcule les expressions suivantes.
\(4^{2} + 6 \cdot 2^{-1} =\)
\(3^{-2} + 8 + 5^{-1} =\)
\(7^{3} \cdot 3^{-3} + 6^{2} - 9^{-1} =\)
\(2^{4} + 2^{-4} =\)
Réponses :
\(19\)
\(\frac{374}{45} \approx 8,31\)
\(\frac{1312}{27} \approx 48,59\)
\(\frac{257}{16} = 16,0625\)
\[4^{2} = 4 \times 4 = 16\]
Un exposant négatif signifie que l’on prend l’inverse du nombre :
\[2^{-1} = \frac{1}{2}\]
\[6 \cdot 2^{-1} = 6 \times \frac{1}{2} = 3\]
\[16 + 3 = 19\]
\[4^{2} + 6 \cdot 2^{-1} = 19\]
\[3^{-2} = \frac{1}{3^{2}} = \frac{1}{9}\]
\[5^{-1} = \frac{1}{5}\]
\[\frac{1}{9} + 8 + \frac{1}{5}\]
Le dénominateur commun entre 9 et 5 est 45.
\[\frac{1}{9} = \frac{5}{45}\]
\[\frac{1}{5} = \frac{9}{45}\]
\[\frac{5}{45} + 8 + \frac{9}{45} = \frac{14}{45} + 8\]
\[8 = \frac{360}{45}\]
\[\frac{14}{45} + \frac{360}{45} = \frac{374}{45}\]
\[3^{-2} + 8 + 5^{-1} = \frac{374}{45} \approx 8,31\]
\[7^{3} = 7 \times 7 \times 7 = 343\]
\[3^{-3} = \frac{1}{3^{3}} = \frac{1}{27}\]
\[343 \times \frac{1}{27} = \frac{343}{27}\]
\[6^{2} = 6 \times 6 = 36\]
\[9^{-1} = \frac{1}{9}\]
\[\frac{343}{27} + 36 - \frac{1}{9}\]
Le dénominateur commun est 27.
\[\frac{1}{9} = \frac{3}{27}\]
\[\frac{343}{27} - \frac{3}{27} = \frac{340}{27}\]
\[36 = \frac{972}{27}\]
\[\frac{340}{27} + \frac{972}{27} = \frac{1312}{27}\]
\[ \frac{1312}{27} \approx 48,59\]
\[7^{3} \cdot 3^{-3} + 6^{2} - 9^{-1} = \frac{1312}{27} \approx 48,59\]
\[2^{4} = 2 \times 2 \times 2 \times 2 = 16\]
\[2^{-4} = \frac{1}{2^{4}} = \frac{1}{16}\]
\[16 + \frac{1}{16}\]
\[16 = \frac{256}{16}\]
\[\frac{256}{16} + \frac{1}{16} = \frac{257}{16}\]
\[ \frac{257}{16} = 16,0625\]
\[2^{4} + 2^{-4} = \frac{257}{16} = 16,0625\]