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解：由题意得$\begin{cases}4 - x\geq0 \\x - 4\geq0\end{cases}，$解得$x = 4，$此时$y = 3。$
$\therefore$原式$=(a + b)^{2}-ab$
$=(\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y})^{2}-(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})$
$=(2\sqrt{x})^{2}-(x - y)=4x - x + y=3x + y=3\times4 + 3$
$=12 + 3=15。$
B
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8
解：$\because x=\frac{1-\sqrt{2022}}{2}，$$\therefore1 - 2x=\sqrt{2022}。$
$\therefore(1 - 2x)^{2}=1 - 4x + 4x^{2}=2022。$
$\therefore4x^{2}-4x=2021。$
$\therefore$原式$=[(4x^{3}-4x^{2})+(4x^{2}-4x)-2021x - 2022]^{3}$
$=(2021x + 2021-2021x - 2022)^{3}=(-1)^{3}=-1。$
解：因为$(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}) = 1，$所以$a=\sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}。$
同理，$b=\frac{1}{2+\sqrt{3}}，$$c=\frac{1}{\sqrt{5}+2}。$
因为$\sqrt{3}+\sqrt{2}＜2+\sqrt{3}＜\sqrt{5}+2，$
当分子相同时，分母大的分数反而小，所以$a＞b＞c。$
(1)解：$\frac{1}{2-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}=\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=2+\sqrt{3}+\sqrt{3}+\sqrt{2}=2 + 2\sqrt{3}+\sqrt{2}。$
(2)解：$\sqrt{2026}-\sqrt{2025}＜\sqrt{2025}-\sqrt{2024}。$
理由：$\because\sqrt{2026}-\sqrt{2025}=\frac{1}{\sqrt{2026}+\sqrt{2025}}，$
$\sqrt{2025}-\sqrt{2024}=\frac{1}{\sqrt{2025}+\sqrt{2024}}，$
又$\because\sqrt{2026}+\sqrt{2025}＞\sqrt{2025}+\sqrt{2024}，$
$\therefore\frac{1}{\sqrt{2026}+\sqrt{2025}}＜\frac{1}{\sqrt{2025}+\sqrt{2024}}。$
$\therefore\sqrt{2026}-\sqrt{2025}＜\sqrt{2025}-\sqrt{2024}。$
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