Question

Problema e mai jos, in poza. Rog rezolvare completa!

Answer 1

$\Big(\sqrt{3+2\sqrt 2}\Big)^x-\Big(\sqrt{3-2\sqrt 2}\Big)^x = \dfrac{3}{2}\\ \\ \Rightarrow \Bigg(\sqrt{\dfrac{9-4\cdot 2}{3-2\sqrt 2}}\Bigg)^x - \Big(\sqrt{3-2\sqrt 2}\Big)^x = \dfrac{3}{2}\\ \\ \Rightarrow\Bigg(\sqrt{\dfrac{1}{3-2\sqrt 2}}\Bigg)^x - \Big(\sqrt{3-2\sqrt 2}\Big)^x = \dfrac{3}{2}\\ \\ \Rightarrow \dfrac{1}{\Big(\sqrt{3-2\sqrt 2}\Big)^x} - \Big(\sqrt{3-2\sqrt 2}\Big)^x = \dfrac{3}{2} \\ \\ \\ \text{Notez }\Big(\sqrt{3-2\sqrt 2}\Big)^x = t,\quad t>0\\ \\\\ \Rightarrow \dfrac{1}{t}- t = \dfrac{3}{2}\\ \\ \Rightarrow 2-2t^2 = 3t \\ \\ \Rightarrow 2t^2+3t-2 = 0$

$\Delta = 25 \Rightarrow t_{1,2} = \dfrac{-3\pm 5}{4} \Rightarrow t = \dfrac{2}{4}= \dfrac{1}{2}\\ \\ \Rightarrow \Big(\sqrt{3-2\sqrt 2}\Big)^x = \dfrac{1}{2}\\ \\ \Rightarrow \Big(\sqrt{(1-\sqrt 2)^2}\Big)^x = \dfrac{1}{2} \\ \\ \Rightarrow |\sqrt 2 - 1|^x = \dfrac{1}{2}\\ \\ \Rightarrow x = \log_{\sqrt 2 - 1}\dfrac{1}{2}\\ \\ \Rightarrow x = \dfrac{1}{\log_{\dfrac{1}{2}}\sqrt{3-2\sqrt 2}}\\ \\ \Rightarrow x = \dfrac{1}{\dfrac{1}{2}\log_{\dfrac{1}{2}}(3-2\sqrt 2)}$

$\Rightarrow x = \dfrac{2}{\dfrac{\lg(3-2\sqrt 2)}{\lg \dfrac{1}{2}}}\Rightarrow x = \dfrac{2\lg \dfrac{1}{2}}{\lg(3-2\sqrt 2)} \Rightarrow \\ \\ \Rightarrow x = \dfrac{2\lg 2}{-\lg(3-2\sqrt 2)} \Rightarrow x = \dfrac{2\lg 2}{\lg(3-2\sqrt2)^{-1}} \Rightarrow \\ \\ \\ \Rightarrow \boxed{x = \dfrac{2\lg 2}{\lg(3+2\sqrt 2)}}$

$\\\Rightarrow\, c)\,\,\text{corect}$

Answer 2

Explicație pas cu pas:

................................