$a)$ ĐKXĐ: $x > 0$ 

$D = \dfrac{\sqrt{x}}{x + \sqrt{x}} : \left( \dfrac{1}{\sqrt{x}} + \dfrac{\sqrt{x}}{\sqrt{x} + 1} \right)$

$= \dfrac{\sqrt{x}}{\sqrt{x}(\sqrt{x} + 1)} : \dfrac{\sqrt{x} + 1 + (\sqrt{x})^2}{\sqrt{x}(\sqrt{x} + 1)}$

$= \dfrac{1}{\sqrt{x} + 1} : \dfrac{x + \sqrt{x} + 1}{\sqrt{x}(\sqrt{x} + 1)}$

$= \dfrac{1}{\sqrt{x} + 1} \cdot \dfrac{\sqrt{x}(\sqrt{x} + 1)}{x + \sqrt{x} + 1}$

$= \dfrac{\sqrt{x}}{x + \sqrt{x} + 1}$

Vậy $D = \dfrac{\sqrt{x}}{x + \sqrt{x} + 1}$

$b)$ Để $D = \dfrac{2}{7}$ thì:

$\dfrac{\sqrt{x}}{x + \sqrt{x} + 1} = \dfrac{2}{7}$

$7\sqrt{x} = 2(x + \sqrt{x} + 1)$

$7\sqrt{x} = 2x + 2\sqrt{x} + 2$

$2x - 5\sqrt{x} + 2 = 0$

$2x - 4\sqrt{x} - \sqrt{x} + 2 = 0$

$2\sqrt{x}(\sqrt{x} - 2) - (\sqrt{x} - 2) = 0$

$(2\sqrt{x} - 1)(\sqrt{x} - 2) = 0$

$\left[\begin{aligned} 2\sqrt{x} - 1 &= 0 \\ \sqrt{x} - 2 &= 0 \end{aligned}\right.$

$\left[\begin{aligned} \sqrt{x} &= \dfrac{1}{2} \\ \sqrt{x} &= 2 \end{aligned}\right.$

$\left[\begin{aligned} x &= \dfrac{1}{4} \text{ (tmđk)} \\ x &= 4 \text{ (tmđk)} \end{aligned}\right.$

Vậy $x = \dfrac{1}{4}$ hoặc $x = 4$ thì $D = \dfrac{2}{7}$