Toán Lớp 12: Tìm TXĐ của hs:
y =\sqrt{log_2 (x²+2) . log_{2-x} 2 -2}
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Toán Lớp 12: Tìm TXĐ của hs:
y =\sqrt{log_2 (x²+2) . log_{2-x} 2 -2}
Comments ( 1 )
\quad y = \sqrt{\log_{\displaystyle 2}(x^2 + 2).\log_{\displaystyle{2-x}}2 – 2}\\
\text{Hàm số xác định khi và chỉ khi}\\
\quad \begin{cases}0 < 2- x \ne 1\\\log_{\displaystyle 2}(x^2 + 2).\log_{\displaystyle{2-x}}2 – 2\geqslant 0\end{cases}\\
\Leftrightarrow \begin{cases}x < 2\\x\ne 1\\\log_{\displaystyle 2}(x^2 + 2)\cdot \dfrac{1}{\log_{\displaystyle{2}}(2-x)}-2\geqslant 0\qquad (*)\end{cases}\\
(*)\Leftrightarrow \dfrac{\log_{\displaystyle 2}(x^2 + 2)}{\log_{\displaystyle{2}}(2-x)} – 2 \geqslant 0\\
\Leftrightarrow \dfrac{\log_{\displaystyle 2}(x^2 + 2) – 2\log_{\displaystyle{2}}(2-x)}{\log_{\displaystyle{2}}(2-x)}\geqslant 0\\
\Leftrightarrow \dfrac{\log_{\displaystyle 2}\dfrac{x^2 + 2}{(x-2)^2}}{\log_{\displaystyle{2}}(2-x)}\geqslant 0\\
\Leftrightarrow \left[\begin{array}{l}\begin{cases}\log_{\displaystyle 2}\dfrac{x^2 + 2}{(x-2)^2} \geqslant 0\\\log_{\displaystyle{2}}(2-x)>0\end{cases}\\\begin{cases}\log_{\displaystyle 2}\dfrac{x^2 + 2}{(x-2)^2} \leqslant 0\\\log_{\displaystyle{2}}(2-x) <0\end{cases}\end{array}\right.\\
\Leftrightarrow \dfrac12 \leqslant x < 1\\
\text{Vậy}\ TXD: D = \left[\dfrac12;1\right)
\end{array}\)