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解​$：(1)$​原式​$=\frac {3-2\sqrt {3}+1}{4}$​
​$=\frac {4-2\sqrt {3}}{4}$​
​$= \frac {2 - \sqrt {3}}{2}$​
解​$：(2)$​原式​$=\frac {1+2\sqrt {5}+5}{4}$​
​$=\frac {6+2\sqrt {5}}{4}$​
​$= \frac {3 + \sqrt {5}}{2}$​
解:​$S=\frac {1}{2}(4+\sqrt {3})(4-\sqrt {3})$​
​$=\frac {1}{2}(16-3)$​
​$=\frac {13}{2}$​
答：该菱形面积为​$\frac {13}{2}$​。
解​$：(1)x^2+2xy + y^2=(x + y)^2$​
​$=(\sqrt {3}+1+\sqrt {3}-1)^2$​
​$=(2\sqrt {3})^2$​
​$=12$​
​$(2)x^2-y^2=(x + y)(x - y)$​
​$=2\sqrt {3}×2$​
​$=4\sqrt {3}$​
解:​$c^2=(2\sqrt {3}+1)^2+(2\sqrt {3}-1)^2$​
​$=(12 + 4\sqrt {3}+1)+(12 - 4\sqrt {3}+1)$​
​$=26$​
​$c=\sqrt {26}$​
答：斜边长为​$\sqrt {26}$​。
解​$： (1)① S_{2} - S_{1}=(1 + 2\sqrt {2})^2-(1 + \sqrt {2})^2$​
​$=(1 + 4\sqrt {2} + 8)-(1 + 2\sqrt {2} + 2)=9 + 4\sqrt {2}-3 - 2\sqrt {2}=6 + 2\sqrt {2}$​
​$② S_{3} - S_{2}=(1 + 3\sqrt {2})^2-(1 + 2\sqrt {2})^2$​
​$=(1 + 6\sqrt {2} + 18)-(1 + 4\sqrt {2} + 8)=19 + 6\sqrt {2}-9 - 4\sqrt {2}=10 + 2\sqrt {2}$​
​$③ S_{4} - S_{3}=(1 + 4\sqrt {2})^2-(1 + 3\sqrt {2})^2$​
​$=(1 + 8\sqrt {2} + 32)-(1 + 6\sqrt {2} + 18)=33 + 8\sqrt {2}-19 - 6\sqrt {2}=14 + 2\sqrt {2}$​
​$(2)$​猜想​$S_{n+1}-S_{n}=4n + 2 + 2\sqrt {2}，$​理由：
​$S_{n}=(1 + n\sqrt {2})^2=2n^2 + 2\sqrt {2}n + 1，$​
​$S_{n+1}=(1 + (n+1)\sqrt {2})^2=2(n+1)^2 + 2\sqrt {2}(n+1)+1=2n^2 + 4n + 2 + 2\sqrt {2}n + 2\sqrt {2} + 1，$​
则​$S_{n+1}-S_{n}=(2n^2 + 4n + 2 + 2\sqrt {2}n + 2\sqrt {2} + 1)-(2n^2 + 2\sqrt {2}n + 1)=4n + 2 + 2\sqrt {2}$​
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