Exercițiul 5 cu demonstrarea răspunsului, vă rog. Ofer coroană și 60 de puncte
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Salut,
![a=\dfrac{1}{log_{(\sqrt3+1)}{(\sqrt2-1)}}+\dfrac{1}{log_{(\sqrt3+1)}{(\sqrt2+1)}}=\dfrac{log_{(\sqrt3+1)}{(\sqrt2-1)}+log_{(\sqrt3+1)}{(\sqrt2+1)}}{log_{(\sqrt3+1)}{(\sqrt2-1)}\cdot log_{(\sqrt3+1)}{(\sqrt2+1)}}=\\\\=\dfrac{log_{(\sqrt3+1)}{[(\sqrt2-1)}\cdot(\sqrt2+1)]}{log_{(\sqrt3+1)}{(\sqrt2-1)}\cdot log_{(\sqrt3+1)}{(\sqrt2+1)}}=\dfrac{log_{(\sqrt3+1)}{[(\sqrt2)^2-1^2]}}{log_{(\sqrt3+1)}{(\sqrt2-1)}\cdot log_{(\sqrt3+1)}{(\sqrt2+1)}}=\\\\=\dfrac{log_{(\sqrt3+1)}1}{log_{(\sqrt3+1)}{(\sqrt2-1)}\cdot log_{(\sqrt3+1)}{(\sqrt2+1)}}=0,\ deci\ a=0.\\\\\\log_{\frac{1}{2}}\dfrac{1}{512}=log_{\frac{1}{2}}\left(\dfrac{1}{2}\right)^9=9.\ log_39=log_33^2=2.\ log_42=\dfrac{1}{log_24}=\\\\=\dfrac{1}{log_22^2}=\dfrac{1}2,\ deci\ b=\dfrac{1}2\Rightarrow b^2=\dfrac{1}4,\ deci\ 4b^2=1\Rightarrow a^2+4b^2=1.](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B1%7D%7Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2-1%29%7D%7D%2B%5Cdfrac%7B1%7D%7Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2%2B1%29%7D%7D%3D%5Cdfrac%7Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2-1%29%7D%2Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2%2B1%29%7D%7D%7Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2-1%29%7D%5Ccdot+log_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2%2B1%29%7D%7D%3D%5C%5C%5C%5C%3D%5Cdfrac%7Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%5B%28%5Csqrt2-1%29%7D%5Ccdot%28%5Csqrt2%2B1%29%5D%7D%7Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2-1%29%7D%5Ccdot+log_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2%2B1%29%7D%7D%3D%5Cdfrac%7Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%5B%28%5Csqrt2%29%5E2-1%5E2%5D%7D%7D%7Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2-1%29%7D%5Ccdot+log_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2%2B1%29%7D%7D%3D%5C%5C%5C%5C%3D%5Cdfrac%7Blog_%7B%28%5Csqrt3%2B1%29%7D1%7D%7Blog_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2-1%29%7D%5Ccdot+log_%7B%28%5Csqrt3%2B1%29%7D%7B%28%5Csqrt2%2B1%29%7D%7D%3D0%2C%5C+deci%5C+a%3D0.%5C%5C%5C%5C%5C%5Clog_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5Cdfrac%7B1%7D%7B512%7D%3Dlog_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E9%3D9.%5C+log_39%3Dlog_33%5E2%3D2.%5C+log_42%3D%5Cdfrac%7B1%7D%7Blog_24%7D%3D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7Blog_22%5E2%7D%3D%5Cdfrac%7B1%7D2%2C%5C+deci%5C+b%3D%5Cdfrac%7B1%7D2%5CRightarrow+b%5E2%3D%5Cdfrac%7B1%7D4%2C%5C+deci%5C+4b%5E2%3D1%5CRightarrow+a%5E2%2B4b%5E2%3D1. "a=\dfrac{1}{log_{(\sqrt3+1)}{(\sqrt2-1)}}+\dfrac{1}{log_{(\sqrt3+1)}{(\sqrt2+1)}}=\dfrac{log_{(\sqrt3+1)}{(\sqrt2-1)}+log_{(\sqrt3+1)}{(\sqrt2+1)}}{log_{(\sqrt3+1)}{(\sqrt2-1)}\cdot log_{(\sqrt3+1)}{(\sqrt2+1)}}=\\\\=\dfrac{log_{(\sqrt3+1)}{[(\sqrt2-1)}\cdot(\sqrt2+1)]}{log_{(\sqrt3+1)}{(\sqrt2-1)}\cdot log_{(\sqrt3+1)}{(\sqrt2+1)}}=\dfrac{log_{(\sqrt3+1)}{[(\sqrt2)^2-1^2]}}{log_{(\sqrt3+1)}{(\sqrt2-1)}\cdot log_{(\sqrt3+1)}{(\sqrt2+1)}}=\\\\=\dfrac{log_{(\sqrt3+1)}1}{log_{(\sqrt3+1)}{(\sqrt2-1)}\cdot log_{(\sqrt3+1)}{(\sqrt2+1)}}=0,\ deci\ a=0.\\\\\\log_{\frac{1}{2}}\dfrac{1}{512}=log_{\frac{1}{2}}\left(\dfrac{1}{2}\right)^9=9.\ log_39=log_33^2=2.\ log_42=\dfrac{1}{log_24}=\\\\=\dfrac{1}{log_22^2}=\dfrac{1}2,\ deci\ b=\dfrac{1}2\Rightarrow b^2=\dfrac{1}4,\ deci\ 4b^2=1\Rightarrow a^2+4b^2=1.")
Green eyes.
Green eyes.
ancutadraguta2:
Vă mulțumesc din nou! c:
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