a,Ta có :  \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)

Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

b, Đặt A =  \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)

\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)

Vậy (*) = 0 

1: 

Ta có: \(\sqrt{2}-\sqrt{6}\)

\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)

\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

Ta có: \(X=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)

<=> \(X^2=6-3\sqrt{2+\sqrt{3}}+2+\sqrt{2+\sqrt{3}}-2\sqrt{3}.\sqrt{4-\left(2+\sqrt{3}\right)}\)

<= \(X^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3}.\sqrt{2-\sqrt{3}}\)

<=> \(X^2=8-\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{6}\left(\sqrt{3}-1\right)\)

<=> \(X^2=8-4\sqrt{2}\)

<=> \(X^2-8=-4\sqrt{2}\)

=> \(X^4-16X+64=32\)

<=> \(X^4-16X^2+32=0\)

Vậy X là nghiệm phương trình \(X^4-16X^2+32=0\)

\(\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{2-\sqrt{3}}{4-3}+\dfrac{2+\sqrt{3}}{4-3}=2-\sqrt{3}+2+\sqrt{3}=4\)

a: \(\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac14\cdot\sqrt8\right)\cdot3\sqrt6\)

\(=\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac12\sqrt2\right)\cdot3\sqrt6\)

\(=\left(2\sqrt6-4\sqrt3+\frac92\cdot\sqrt2\right)\cdot3\sqrt6\)

\(=2\sqrt6\cdot3\sqrt6-4\sqrt3\cdot3\sqrt6+\frac92\cdot\sqrt2\cdot3\sqrt6\)

\(=36-12\sqrt{18}+\frac{27}{2}\sqrt{12}=36-36\sqrt2+27\sqrt3\)

b: \(\left(\sqrt{\frac17}-\sqrt{\frac{16}{7}}+\sqrt7\right):\sqrt7=\left(\frac{\sqrt7}{7}-\frac{4\sqrt7}{7}+\sqrt7\right):\sqrt7\)

\(=\frac17-\frac47+1=\frac87-\frac47=\frac47\)

c: \(\left(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\right)^2\)

\(=3-\sqrt5+3+\sqrt5+2\cdot\sqrt{\left(3-\sqrt5\right)\left(3+\sqrt5\right)}\)

\(=6+2\cdot\sqrt{9-5}=6+2\cdot2=10\)

Trả lời:

\(A=\sqrt{3}-\frac{\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)

\(A=\sqrt{3}+\frac{\sqrt{6}}{\sqrt{2}-1}-\frac{2\sqrt{2}+2}{\sqrt{2}+1}\)

\(A=\sqrt{3}+\frac{\sqrt{6}.\left(\sqrt{2}+1\right)}{2-1}-\frac{2.\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)

\(A=\sqrt{3}+\sqrt{6}.\left(\sqrt{2}+1\right)-2\)

\(A=\sqrt{3}+\sqrt{12}+\sqrt{6}-2\)

\(A=\sqrt{3}+2\sqrt{3}+\sqrt{6}-2\)

\(A=3\sqrt{3}+\sqrt{6}-2\)

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