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Simplification de $\mathcal{A}$
- $\mathcal{A} = 15 \ln(7 + 4\sqrt{3}) + 15 \ln(7 - 4\sqrt{3}) = 15 \ln[(7 + 4\sqrt{3})(7 - 4\sqrt{3})]$
- Or, $(7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 49 - 48 = 1$.
- $\mathcal{A} = 15 \ln(1) = 0$.
- $\mathcal{A} = 0$
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Simplification de $\mathcal{B}$
- $\mathcal{B} = \ln 2 + \ln[(2 + \sqrt{2 + \sqrt{2}})(2 - \sqrt{2 + \sqrt{2}})]$
- $\mathcal{B} = \ln 2 + \ln[4 - (2 + \sqrt{2})] = \ln 2 + \ln(2 - \sqrt{2})$
- $\mathcal{B} = \ln[2(2 - \sqrt{2})]$
- $\mathcal{B} = \ln(4 - 2\sqrt{2})$