$\begin{array}{l}118)\,\,\log_2(x^2 - 3) - \log_2(6x - 10) + 1 = 0\qquad (*)\\ ĐKXĐ: \, \begin{cases}x^2 - 3 > 0\\6x - 10 > 0\end{cases}\Leftrightarrow x > \sqrt3\\\ (*) \Leftrightarrow \log_2\dfrac{x^2 - 3}{6x- 10} = -1\\ \Leftrightarrow \dfrac{x^2 -3}{6x - 10} = 2^{-1}\\ \Leftrightarrow 2(x^2 - 3) = 6x - 10\\ \Leftrightarrow \left[\begin{array}{l}x = 1\quad (loại)\\x = 2\quad (nhận)\end{array}\right.\\ 124)\,\,\dfrac{1}{4 + \log_2x} + \dfrac{1}{2 - \log_2x} = 1\qquad (x >0)\\ Đặt\,\,t = \log_2x\\ \text{Phương trình trở thành:}\\ \dfrac{1}{4 + t} + \dfrac{1}{2 - t} = 1\\ \Leftrightarrow 2 - t + 4 + t = (4+t)(2-t)\\ \Leftrightarrow t^2 + 2t - 2 =0\\ \Leftrightarrow \left[\begin{array}{l}t = -1 - \sqrt3\\t = -1 + \sqrt3\end{array}\right.\\ Ta\,\,được:\\ \quad \left[\begin{array}{l}\log_2x = -1 - \sqrt3\\\log_2x = -1 + \sqrt3\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = 2^{-1 - \sqrt3}\\x = 2^{-1 + \sqrt3}\end{array}\right.\\ 122)\,\,\ln(x+1) + \ln(x+3) = \ln(x+7)\qquad (x > - 1)\\ \Leftrightarrow \ln[(x+1)(x+3)] = \ln(x+7)\\ \Leftrightarrow (x+1)(x+3) = x+7\\ \Leftrightarrow x^2 + 3x - 4 = 0\\ \Leftrightarrow \left[\begin{array}{l}x = 1\quad (nhận)\\x = -4 \quad (loại)\end{array}\right. \end{array}$
