Đáp án:
$a)\quad S =\left\{-\dfrac12\right\}$
$b)\quad S =\left\{\dfrac35;1\right\}$
Giải thích các bước giải:
$a)\quad \sqrt{x^2 - 6x + 9}=\sqrt{16 + 8x + x^2}$
$\Leftrightarrow |x - 3| = |x + 4|$
$\Leftrightarrow \left[\begin{array}{l}x + 4= x - 3\\x + 4 = 3 - x\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}4 = - 3\quad (vl)\\2x = - 1\end{array}\right.$
$\Leftrightarrow x = -\dfrac12$
Vậy $S =\left\{-\dfrac12\right\}$
$b)\quad \sqrt{4 - 12x + 9x^2} - \sqrt{4x^2 - 4x + 1}= 0$
$\Leftrightarrow \sqrt{4 - 12x + 9x^2} = \sqrt{4x^2 - 4x + 1}$
$\Leftrightarrow |2 - 3x| = |2x -1|$
$\Leftrightarrow \left[\begin{array}{l}2 - 3x = 2x - 1\\3x - 2 = 2x -1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}5x = 3\\x= 1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac35\\x= 1\end{array}\right.$
Vậy $S =\left\{\dfrac35;1\right\}$
