Root of the problem

Denominator becomes

$\sqrt{32-2(5-\sqrt{21})} - \sqrt{21}$

$=\sqrt{22+2\sqrt{21}} - \sqrt{21}$

$=\sqrt{(1+\sqrt{21})^2} - \sqrt{21}$

$=1+\sqrt{21} - \sqrt{21}$

$=1$

Numerator becomes

$\sqrt{5-\sqrt{21}}$

$=\sqrt{\dfrac{10-2\sqrt{21}}{2}}$

$=\sqrt{\dfrac{(\sqrt{7}-\sqrt{3})^2}{2}}$

$=\dfrac{(\sqrt{7}-\sqrt{3})}{\sqrt{2}}$

$=\sqrt{\dfrac{7}{2}} - \sqrt{\dfrac{3}{2}}$

Thus $a+b+c+d=7+2+3+2=\boxed{14}$

An educated guess to approach problem with radicals is checking if radicals can be simplified as $\sqrt{A+\sqrt{B}} =$ $\sqrt{(\alpha+\sqrt{\beta})^2} =$ $\alpha+\sqrt{\beta}$ . I first checked on $\sqrt{5-\sqrt{21}}$ . It did not work. So let us check on $\sqrt{32-2(5-\sqrt{21})}$ .

$\begin{aligned} \sqrt{32-2(5-\sqrt{21})} & = \sqrt{(\alpha+\sqrt{\beta})^2} \\ \sqrt{22+2\sqrt{21}} & = \sqrt{\alpha^2 + \beta + 2\alpha \sqrt{\beta}} \end{aligned}$

Equating the rational and irrational parts: $\begin{cases} \alpha^2 + \beta = 22 \\ 2\alpha \sqrt{\beta} = 2\sqrt{21} \end{cases}$

It is reasonable to assume that $\sqrt{\beta} = \sqrt{21}$ or $\beta = 21$ , then $\alpha = 1$ which satisfies the equations. $\implies \sqrt{32-2(5-\sqrt{21})} = 1 + \sqrt{21}$ . Therefore, we have:

$\begin{aligned} X = \frac {\sqrt{5-\sqrt{21}}}{\sqrt{32-2(5-\sqrt{21})} - \sqrt{21}} & = \frac {\sqrt{5-\sqrt{21}}}{1+\sqrt{21} - \sqrt{21}} = \sqrt{5-\sqrt{21}} \end{aligned}$

Now, let $X = \sqrt x - \sqrt y$ .

$\begin{aligned} \implies \sqrt{x} - \sqrt{y} & = \sqrt{5-\sqrt{21}} & \small \color{#3D99F6}{\text{Squaring both sides,}} \\ x + y - 2\sqrt{xy} & = 5 - \sqrt{21} \end{aligned}$

$\implies \begin{cases} x+y = 5 & ...(1) \\ 4xy = 21 & ...(2) \end{cases}$

$\begin{aligned} (2): \quad y & = \frac {21}{4x} \\ (1): \quad x + \frac {21}{4x} & = 5 \\ 4x^2 - 20x + 21 & = 0 \\ (2x-3)(2x-7) & = 0 \\ \implies x & = \frac 72 \\ y & = \frac 32 \end{aligned}$

$\implies X = \sqrt{\dfrac 72} - \sqrt{\dfrac 32} \implies a + b+c+d = 7+2+3+2 = \boxed{14}$