Ta có : \(A=\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{2^2-2.2\sqrt{3}+3}\)

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

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

Ta có : \(B=\sqrt{3+\sqrt{8}+\sqrt{3-\sqrt{8}}}\)

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

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

b,\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)  \(=\sqrt{8\sqrt{3}}-2\sqrt{50\sqrt{3}}+4\sqrt{8\sqrt{3}}\)

\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}\)

\(=0\)

d,\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(\sqrt{2}A=\sqrt{2}(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}})\)

\(\sqrt2A=\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)

\(\sqrt2A=\sqrt{(\sqrt5-1)^2}\) \(+\sqrt{(\sqrt5+1)^2}\)    \(=\sqrt5-1 +\sqrt5+1=2\sqrt5\)

\(\Rightarrow A=\dfrac{2\sqrt5}{\sqrt2}\) \(=\sqrt{10}\)

a. \(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)

\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\) 

\(=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{3\sqrt{5}-3+5-\sqrt{5}}{2\left(\sqrt{5}+1\right)}\)  

\(=\frac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=\frac{2\left(\sqrt{5}+1\right)}{2\left(\sqrt{5}+1\right)}=1\)

a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=1\)

b: \(=\sqrt{\sqrt{3}}\left(2\sqrt{2}-2\cdot5\sqrt{2}+4\cdot8\sqrt{2}\right)\)

\(=\sqrt{\sqrt{3}}\cdot24\sqrt{2}\)

d: \(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

Lời giải:

$A=\sqrt{7}-(4\sqrt{3}+2\sqrt{3})+\sqrt{4}$

$=\sqrt{7}-6\sqrt{3}+2$

$B=\frac{7-\sqrt{5}}{\sqrt{5}-1}.\frac{-3+\sqrt{5}}{\sqrt{5}+1}$

$=\frac{(7-\sqrt{5})(-3+\sqrt{5})}{(\sqrt{5}-1)(\sqrt{5}+1)}$

$=\frac{-26+10\sqrt{5}}{4}=\frac{-13+5\sqrt{5}}{2}$

$C=(\sqrt{3}+1).\frac{\sqrt{14}-6\sqrt{3}}{\sqrt{5}+1}$

$=\frac{-18-6\sqrt{3}+\sqrt{14}+\sqrt{42}}{\sqrt{5}+1}$

Với biểu thức của bạn thì càng cố rút gọn thì càng dài. Bạn cần gấp nhưng đề bài lại viết sai?

c/ = \(\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{25+2.3.5.\sqrt{2}+18}\)

\(=5+3\sqrt{2}\)

d/ \(=\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)

\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)

\(=\sqrt{13+6\left(\sqrt{3}+1\right)}\)

\(=\sqrt{19+6\sqrt{2}}\)

\(=3\sqrt{2}+1\)