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$ 5\sqrt{\frac{1}{6}}$

$ 6\sqrt{\frac{1}{7}}$

$解：???(2)???原式???=\sqrt {\frac {13×15+1}{15}}=\sqrt {\frac {142}{15}}=14\sqrt {\frac {1}{15}}???$

$??? (3)\sqrt {n+\frac {1}{n+2}}=(n+1)\sqrt {\frac {1}{n+2}}(n≥1)??$

?

2

$解：原式???={[(\sqrt {6}-\sqrt {5})×(\sqrt {6}+\sqrt {5})]}^{2021}???$

$??? ×(\sqrt {6}+\sqrt {5})???$

$??? =1×(\sqrt {6}+\sqrt {5})???$

$??? =\sqrt {6}+\sqrt {5}$

$解：原式???=(1+\sqrt {3}-\sqrt {2})2???-(1+\sqrt {3}+\sqrt {2})2???$

$??? =-2\sqrt {2}×(2+2\sqrt {3})???$

$??? =-4\sqrt {2}-4\sqrt {6}$

$解：??? (1)???若???a???有意義，則???8-x≥0，??????x≤8???$

$ 若???b???有意義，則???3x+4≥0，??????x≥-\frac {4}{3}???$

$ 若???c???有意義，則???x+2≥0，??????x≥-2???$

$ 當(dāng)???-\frac {4}{3}≤x≤8???時(shí)，???a、??????b、??????c???都有意義$

$??? (2)???若???a、??????b、??????c???為直角三角形的三邊，則???-\frac {4}{3}\lt x\lt 8???$

$??? ①a2+b2= c2???時(shí)，???(8-x)+ (3x+4)=x+2???$

$??? x=-10，???不滿足???-\frac {4}{3}\lt x\lt 8???$

故此時(shí)不成立

$??? ②a2+c2= b2???時(shí)，???(8-x)+(x+2)= 3x+4???$

$??? x=2，???滿足???-\frac {4}{3}\lt x\lt 8???$

$??? ③c2+b2=a2???時(shí)，???(3x+4)+(x+2)=8-x???$

$??? x=\frac {2}{5}，???滿足???-\frac {4}{3}\lt x\lt 8???$

$ 綜上所述：當(dāng)???x= 2???或???\frac {2}{5}???時(shí)，???a、??????b、??????c???為直角三角形的三邊。$

21

4

$??2+\sqrt 3??$

$??\text {解: (3) 原式 }=\sqrt{(\sqrt{2})^{2}-2 \sqrt{2}+1}+ ????\sqrt {(\sqrt {3})^2-2 ×\sqrt {3} ×\sqrt {2}+(\sqrt {2})^2}+ ??$

$??\sqrt {2^2-2 ×2 ×\sqrt {3}+(\sqrt {3})^2} +···+????\sqrt {(\sqrt {n+1})^2+2 \sqrt {n+1} · \sqrt {n}+(\sqrt {n})^2} ??$

$??=\sqrt {(\sqrt {2}-1)^2}+\sqrt {(\sqrt {3}-\sqrt {2})^2}+\sqrt {(2-\sqrt {3})^2} ??+···+\sqrt {(\sqrt {n+1}-\sqrt {n})^2} ??$

$??=\sqrt {2}-1+\sqrt {3}-\sqrt {2}+2-\sqrt {3}+···+\sqrt {n+1}-\sqrt {n} ??$

$??=\sqrt {n+1}-1 .??$

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