Đáp án:
$x = \dfrac{1}{k^2}\quad \forall k \in \Bbb Z^+;\, k \ne 1$
Giải thích các bước giải:
$B =\left(\dfrac{1+\sqrt x}{x -2\sqrt x} -\dfrac{1}{\sqrt x -2}\right):\dfrac{1}{x -3\sqrt x +2}\qquad (x > 0;\, x \ne 1;\, x \ne 2)$
$\to B =\left(\dfrac{1+\sqrt x}{\sqrt x(\sqrt x -2)} -\dfrac{\sqrt x}{\sqrt x(\sqrt x -2)}\right):\dfrac{1}{(\sqrt x -1)(\sqrt x -2)}$
$\to B =\dfrac{1 +\sqrt x -\sqrt x}{\sqrt x(\sqrt x -2)}\cdot (\sqrt x-1)(\sqrt x-2)$
$\to B = \dfrac{\sqrt x -1}{\sqrt x}$
$\to B = 1 -\dfrac{1}{\sqrt x}$
$B = 1 -\dfrac{1}{\sqrt x}\in\Bbb Z$
$\to \dfrac{1}{\sqrt x}\in \Bbb Z$
$\to \sqrt{\dfrac{1}{ x}} \in \Bbb Z$
$\to \sqrt{\dfrac{1}{x}} = k \in \Bbb Z$
Do $\sqrt{\dfrac1x} >0$
nên $k \in \Bbb Z^+$
Khi đó:
$\dfrac1x = k^2$
$\to x = \dfrac{1}{k^2}$
Ta lại có:
$+) \quad x \ne 1 \to \dfrac{1}{k^2} \ne 1 \to k \ne \pm 1$
mà $k\in \Bbb Z^+$
nên $k\ne 1$
$+) \quad x \ne 2 \to \dfrac{1}{k^2} \ne 2 \to k \ne \pm \sqrt2$
mà $k\in \Bbb Z^+$
nên $x \ne 2 \quad \forall k \in \Bbb Z^+$
Vậy $x = \dfrac{1}{k^2}\quad \forall k \in \Bbb Z^+;\, k \ne 1$
