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解：∵二次根式​$\sqrt {x^2-4}$​，​$\sqrt {8-2x^2}$​有意义，
∴​$\begin {cases}x^2-4≥0，\\8-2x^2≥0，\end {cases}$​
∴​$x^2=4，$​
∴​$x=\pm 2.$​
又∵分母​$x+2≠0$​，
∴​$x=2$​，此时​$y=\frac {0+0+12}{2+2}=3$​，
∴原式​$=\sqrt {\frac {3}{2}}+\sqrt {\frac {2}{3}}=\frac {\sqrt {6}}{2}+\frac {\sqrt {6}}{3}=\frac {5\sqrt {6}}{6}$​
​$ B$​
$8$
$8$
解：∵​$x=\frac {1-\sqrt {2022}}{2}$​，
∴​$1-2x=\sqrt {2022}.$​
∴​$(1-2x)^2=1-4x+4x^2=2022.$​
∴​$4x^2-4x=2021.$​
∴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=2-\sqrt {3}=\frac {1}{2+\sqrt {3}}$​，​$c=\sqrt {5}-2=\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$​
解：​$(1) \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) \sqrt {2026}-\sqrt {2025}＜\sqrt {2025}-\sqrt {2024}.$​
理由：∵​$\sqrt {2026}-\sqrt {2025}=\frac {1}{\sqrt {2026}+\sqrt {2025}}$​，
​$ \sqrt {2025}-\sqrt {2024}=\frac {1}{\sqrt {2025}+\sqrt {2024}}$​，
又∵​$\sqrt {2026}+\sqrt {2025}＞\sqrt {2025}+\sqrt {2024}$​，
∴​$\frac {1}{\sqrt {2026}+\sqrt {2025}}＜\frac {1}{\sqrt {2025}+\sqrt {2024}}.$​
∴​$\sqrt {2026}-\sqrt {2025}＜\sqrt {2025}-\sqrt {2024}$​
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