📝 Exercices de la série▼
Exercice 1Exercice 2Exercice 3Exercice 4Exercice 5Exercice 6Exercice 7Exercice 8Exercice 9Exercice 10Exercice 11Exercice 12Exercice 13

Exercice 10

  1. Calculer le nombre : A=(32×115)−2(34×112)3×331532×11−1A=\dfrac{\left(3^2\times11^5\right)^{-2}}{\left(3^4\times11^2\right)^3}\times \dfrac{33^{15}}{3^2\times11^{-1}}

  2. Sachant que : b4=1681b^4=\dfrac{16}{81} et b6=64729b^6=\dfrac{64}{729} déterminer b2b^2 sans calculer bb

A=(32)−2×(115)−2(34)3×(112)3×315×111532×11−1=3−4×11−10312×116×315×111532×11−1=311×115314×115=311−14=3−3=133=127\begin{align*} A &= \frac{\left(3^2\right)^{-2} \times \left({11}^5\right)^{-2}}{\left(3^4\right)^3 \times \left({11}^2\right)^3} \times \frac{3^{15} \times {11}^{15}}{3^2 \times {11}^{-1}} \\ &= \frac{3^{-4} \times {11}^{-10}}{3^{12} \times {11}^6} \times \frac{3^{15} \times {11}^{15}}{3^2 \times {11}^{-1}} \\ &= \frac{3^{11} \times {11}^5}{3^{14} \times {11}^5} \\ &= 3^{11 - 14} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27} \end{align*}
b6b4=647291681=64729×8116=4×16×819×81×16=49\begin{align*} \dfrac{b^6}{b^4}=\dfrac{\dfrac{64}{729}}{\dfrac{16}{81}} =\dfrac{64}{729}\times\dfrac{81}{16}=\dfrac{4\times16\times81}{9\times81\times16}=\dfrac{4}{9} \end{align*}

et on a : b6b4=b6−4=b2\dfrac{b^6}{b^4}=b^{6-4}=b^2,

Donc b2=49b^2=\dfrac{4}{9}