Đáp án:
1) $S =\left\{\dfrac12;4\right\}$
2) $\Delta: \dfrac{x-1}{-1}=\dfrac{y-1}{2}=\dfrac{z-3}{2}$
Giải thích các bước giải:
1) $3^{\displaystyle{\log_{\cos\tfrac{\pi}{3}}x - 2\log_x\cos\tfrac{\pi}{3} + 1}} = 2^{\displaystyle{\log_{\sqrt x}\sqrt x +1}}\qquad (ĐK: x > 0)$
$\Leftrightarrow 3^{\displaystyle{\log_{\tfrac12}x - 2\log_x\dfrac12 - 1}} = 2^{\displaystyle{\log_{x^{\tfrac12}}x^{\tfrac12} -1}}$
$\Leftrightarrow 3^{\displaystyle{-\log_{2}x + 2\log_x2 + 1}} = 2^{\displaystyle{\log_xx -1}}$
$\Leftrightarrow 3^{\displaystyle{-\log_{2}x + \dfrac{2}{\log_2x} + 1}} = 2^{\displaystyle{1 -1}}$
$\Leftrightarrow 3^{\displaystyle{-\log_{2}x + \dfrac{2}{\log_2x} + 1}} = 1$
$\Leftrightarrow -\log_{2}x + \dfrac{2}{\log_2x} + 1 = 0$
$\Leftrightarrow - \log_2^2x + \log_2x + 2 = 0$
$\Leftrightarrow \left[\begin{array}{l}\log_2x = -1\\\log_2x = 2\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac12\\x = 4\end{array}\right.$
Vậy $S =\left\{\dfrac12;4\right\}$
2) $(\alpha): 2x - y + 2z - 3 = 0$
$d_1:\dfrac{x-4}{2}=\dfrac{y-1}{2}=\dfrac{z}{-1}$
$d_2:\dfrac{x +3}{2}=\dfrac{y+5}{3}=\dfrac{z-7}{-2}$
Ta có:
$\Delta//(\alpha)$
$\Rightarrow \Delta$ nhận VTPT $\overrightarrow{n}=(2;-1;2)$ của $(\alpha)$ làm VTPT
Ta lại có:
$\quad \begin{cases}\Delta\cap d_1=\{M\}\\\Delta\cap d_2=\{N\}\end{cases}$
$\Rightarrow \begin{cases}M(4+2t_1;1+2t_1;-t_1)\\N(-3+2t_2;-5+3t_2;7-2t_2)\\M, \ N\in \Delta\end{cases}$
$\Rightarrow \overrightarrow{MN}=(-7+2t_2-2t_1;-6+3t_2-2t_1;7-2t_2+t_1)$ là VTCP của $\Delta$
Do đó:
$\quad \overrightarrow{n}.\overrightarrow{MN}= 0$
$\Leftrightarrow 2(-7+2t_2-2t_1) -(-6+3t_2-2t_1)+ 2(7-2t_2+t_1)= 0$
$\Leftrightarrow -3t_2 + 6= 0$
$\Leftrightarrow t_2 = 2$
$\Rightarrow N(1;1;3)$
Mặt khác:
$\quad MN = 3$
$\Rightarrow MN^2 = 9$
$\Leftrightarrow (-7+2t_2-2t_1)^2 + (-6+3t_2-2t_1)^2 + (7-2t_2+t_1)^2 = 9$
$\Leftrightarrow (3+2t_1)^2 + 4t_1^2 + (3+t_1)^2 = 9$
$\Leftrightarrow t_1^2 + 2t_1 + 1 = 0$
$\Leftrightarrow t_1 = -1$
$\Rightarrow \overrightarrow{MN}=(-1;2;2)$
Phương trình đường thẳng $\Delta$ đi qua $N(1;1;3)$ và nhận $\overrightarrow{MN}=(-1;2;2)$ làm VTCP có dạng:
$\Delta: \dfrac{x-1}{-1}=\dfrac{y-1}{2}=\dfrac{z-3}{2}$
