Пусть
По теореме Виета
1)
2)
Ответ: третий вариант
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Verified answer
Пусть![\log_{\sqrt{2}}{x}=t \log_{\sqrt{2}}{x}=t](https://tex.z-dn.net/?f=%5Clog_%7B%5Csqrt%7B2%7D%7D%7Bx%7D%3Dt)
По теореме Виета![\left \{ {{t_{1}+t_{2}=-3} \atop {t_{1}t_{2}=-10}} \right. \Rightarrow t=-5;2 \left \{ {{t_{1}+t_{2}=-3} \atop {t_{1}t_{2}=-10}} \right. \Rightarrow t=-5;2](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bt_%7B1%7D%2Bt_%7B2%7D%3D-3%7D%20%5Catop%20%7Bt_%7B1%7Dt_%7B2%7D%3D-10%7D%7D%20%5Cright.%20%5CRightarrow%20t%3D-5%3B2)
1)![\log_{\sqrt{2}}{x}=2\log_{2}{x}=2\Leftrightarrow\log_{2}{x}=1\Rightarrow x=2 \log_{\sqrt{2}}{x}=2\log_{2}{x}=2\Leftrightarrow\log_{2}{x}=1\Rightarrow x=2](https://tex.z-dn.net/?f=%5Clog_%7B%5Csqrt%7B2%7D%7D%7Bx%7D%3D2%5Clog_%7B2%7D%7Bx%7D%3D2%5CLeftrightarrow%5Clog_%7B2%7D%7Bx%7D%3D1%5CRightarrow%20x%3D2)
2)![2\log_{2}{x}=-5\Leftrightarrow \log_{2}{x}=-\frac{5}{2}\Leftrightarrow x=2^{-\frac{5}{2}}=\frac{1}{\sqrt{2^5}}=\frac{1}{4\sqrt{2}} 2\log_{2}{x}=-5\Leftrightarrow \log_{2}{x}=-\frac{5}{2}\Leftrightarrow x=2^{-\frac{5}{2}}=\frac{1}{\sqrt{2^5}}=\frac{1}{4\sqrt{2}}](https://tex.z-dn.net/?f=2%5Clog_%7B2%7D%7Bx%7D%3D-5%5CLeftrightarrow%20%5Clog_%7B2%7D%7Bx%7D%3D-%5Cfrac%7B5%7D%7B2%7D%5CLeftrightarrow%20x%3D2%5E%7B-%5Cfrac%7B5%7D%7B2%7D%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%5E5%7D%7D%3D%5Cfrac%7B1%7D%7B4%5Csqrt%7B2%7D%7D)
Ответ: третий вариант