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\(\sqrt[3]{8+16x+8\sqrt{3x^2-x-1}}=4\)

Lösungsmenge für x = ?

 06.06.2021
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\(\sqrt[3]{8+16x+8\sqrt{3x^2-x-1}}=4 \ \ \ |^3 \\ 8+16x + 8\sqrt{3x^2-x-1} = 64 \ \ \ |-8-16x \\ 8\sqrt{3x^2-x-1} = -16x+56 \ \ \ |:8 \\ \sqrt{3x^2-x-1} = -2x + 7 \ \ \ |^2 \\ 3x^2-x-1 = 4x^2 -28x +49 \ \ \ |-3x^2+x+1 \\ 0 = x^2 -27x +50 \\ \\ \Rightarrow x_{1/2} = \frac{27 \pm \sqrt{27^2-4\cdot1\cdot 50}}{2} = \frac{27 \pm 23}{2} \Rightarrow x_1 = 2; \ x_2 = 25 \\ \Rightarrow \mathbb{L} = \{ 2, 25 \}\)

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 06.06.2021

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