$2^{x^2 - 3x + 2} = 3^{1-x}$
$\Leftrightarrow 2^{(x -1)(x -2)} = 3^{-(x - 1)}$
$\Leftrightarrow ln2^{(x -1)(x -2)} = ln3^{-(x - 1)}$
$\Leftrightarrow (x-1)(x-2)ln2 + (x-1)ln3= 0$
$\Leftrightarrow (x-1)\left[(x - 2)ln2 + ln3\right] = 0$
$\Leftrightarrow \left[\begin{array}{l}x = 1\\x = -\dfrac{ln3}{ln2} + 2\end{array}\right.$
Do $-\dfrac{ln3}{ln2} + 2 < 1$
nên $\begin{cases}x_1 =-\dfrac{ln3}{ln2} + 2\\x_2 = 1 \end{cases}$
$\Rightarrow S = 4^{-\frac{ln3}{ln2} + 2} + 2^{1 - 2} = \dfrac{41}{18}$
$\Rightarrow$ Không có đáp án
Chỉ tính riêng $4^{-\frac{ln3}{ln2} + 2}$, ta được kết quả $\dfrac{16}{9}$
