Đáp án:
$\dfrac52\leq x \leq 3$
Giải thích các bước giải:
Sửa đề:
$\sqrt{x + 2 +3\sqrt{2x - 5}}+ \sqrt{x -2-\sqrt{2x -5}} = 2\sqrt2\quad \left(x \geq \dfrac52\right)$
$\to \sqrt{2x + 4 + 6\sqrt{2x -5}}+\sqrt{2x - 4 - 2\sqrt{2x - 5}}=4$
$\to \sqrt{2x - 5 + 2.3\sqrt{2x -5} + 9} +\sqrt{2x - 5 - 2\sqrt{2x - 5} +1} = 4$
$\to \sqrt{(\sqrt{2x -5} + 3)^2} +\sqrt{(\sqrt{2x - 5} - 1)^2} = 4$
$\to \sqrt{2x - 5} + 3 +|\sqrt{2x - 5} - 1|= 4$
$\to \left[\begin{array}{l}\sqrt{2x - 5} + 3 +\sqrt{2x - 5} - 1 = 4\quad (x \geq 3)\\\sqrt{2x - 5} + 3 + 1 -\sqrt{2x - 5} = 4\quad \left(\dfrac52\leq x < 3\right)\end{array}\right.$
$\to \left[\begin{array}{l}\sqrt{2x - 5} = 1\\4 = 4\end{array}\right.$
$\to \left[\begin{array}{l}x = 3 \qquad (x \geq 3)\\\text{vô số nghiệm}\quad \left(\dfrac52\leq x < 3\right)\end{array}\right.$
Vậy phương trình có nghiệm $\dfrac52\leq x \leq 3$
