Question

Cum se poate calcula a?

Answer 1

 

$\displaystyle\bf\\a=log_{\sqrt2-1}\Big(\sqrt3+1\Big)+log_{\sqrt2+1}\Big(\sqrt3+1\Big)\\\\a=\frac{1}{log_{\sqrt3+1}\Big(\sqrt2-1\Big)}+\frac{1}{log_{\sqrt3+1}\Big(\sqrt2+1\Big)}\\\\\textbf{Aducem fractiile la acelasi numitor.}$

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$\displaystyle\bf\\a=\frac{log_{\sqrt3+1}\Big(\sqrt2+1\Big)}{\left[log_{\sqrt3+1}\Big(\sqrt2-1\Big)\right]\times\left[log_{\sqrt3+1}\Big(\sqrt2+1\Big)\right]}+\\\\\\+\frac{log_{\sqrt3+1}\Big(\sqrt2-1\Big)}{\left[log_{\sqrt3+1}\Big(\sqrt2-1\Big)\right]\times\left[log_{\sqrt3+1}\Big(\sqrt2+1\Big)\right]}\\\\\\a=\frac{log_{\sqrt3+1}\Big(\sqrt2+1\Big)+log_{\sqrt3+1}\Big(\sqrt2-1\Big)}{\left[log_{\sqrt3+1}\Big(\sqrt2-1\Big)\right]\times\left[log_{\sqrt3+1}\Big(\sqrt2+1\Big)\right]}$

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$\displaystyle\bf\\a=\frac{log_{\sqrt3+1}\left[\Big(\sqrt2+1\Big)\times\Big(\sqrt2-1\Big)\right]}{\left[log_{\sqrt3+1}\Big(\sqrt2-1\Big)\right]\times\left[log_{\sqrt3+1}\Big(\sqrt2+1\Big)\right]}\\\\\\a=\frac{log_{\sqrt3+1}\left[\Big(\sqrt2\Big)^2-1^2\right]}{\left[log_{\sqrt3+1}\Big(\sqrt2-1\Big)\right]\times\left[log_{\sqrt3+1}\Big(\sqrt2+1\Big)\right]}$

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$\displaystyle\bf\\a=\frac{log_{\sqrt3+1}\Big[2-1\Big]}{\left[log_{\sqrt3+1}\Big(\sqrt2-1\Big)\right]\times\left[log_{\sqrt3+1}\Big(\sqrt2+1\Big)\right]}\\\\\\a=\frac{log_{\sqrt3+1}\Big[1\Big]}{\left[log_{\sqrt3+1}\Big(\sqrt2-1\Big)\right]\times\left[log_{\sqrt3+1}\Big(\sqrt2+1\Big)\right]}\\\\\\a=\frac{0}{\left[log_{\sqrt3+1}\Big(\sqrt2-1\Big)\right]\times\left[log_{\sqrt3+1}\Big(\sqrt2+1\Big)\right]}\\\\\\\boxed{\boxed{\bf~a=0}}$