Đáp án:
$\sum x = 1$
Giải thích các bước giải:
$\quad 2^{\displaystyle{x^2+x}} - 4.2^{\displaystyle{x^2 - x}} - 2^{\displaystyle{2x}} + 4 = 0$
$\Leftrightarrow 2^{\displaystyle{2x}}.2^{\displaystyle{x^2-x}}- 4.2^{\displaystyle{x^2 - x}} - 2^{\displaystyle{2x}} + 4 = 0$
$\Leftrightarrow 2^{\displaystyle{x^2 - x}}.\left(2^{\displaystyle{2x}} - 4\right) - \left(2^{\displaystyle{2x}} - 4\right) = 0$
$\Leftrightarrow \left(2^{\displaystyle{2x}} - 4\right)\left(2^{\displaystyle{x^2-x}} - 1\right)= 0$
$\Leftrightarrow \left[\begin{array}{l}2^{\displaystyle{2x}} = 4\\2^{\displaystyle{x^2 - x}} = 1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2x = 2\\x^2 - x = 0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = 1\\x =0\end{array}\right.$
$\to \sum x = 1$
