1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)

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

\(=4\)

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

\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

\(=\left|3-\sqrt{3}\right|+\left|3+\sqrt{3}\right|\)

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

\(=6\)

a)\(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\sqrt{\dfrac{1}{2}\left(16+8\sqrt{3}\right)}-\sqrt{\dfrac{1}{2}\left(16-8\sqrt{3}\right)}\)

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

b)\(=\dfrac{\sqrt{16+2.4\sqrt{5}+5}}{4+\sqrt{5}}.\sqrt{\left(2-\sqrt{5}\right)^2}\)\(=\dfrac{\sqrt{\left(4+\sqrt{5}\right)^2}}{4+\sqrt{5}}\left|2-\sqrt{5}\right|=\sqrt{5}-2\)

a) Ta có: \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)

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

\(=2\sqrt{2}\)

b) Ta có: \(\dfrac{\sqrt{21+8\sqrt{5}}}{4+\sqrt{5}}\cdot\sqrt{9-4\sqrt{5}}\)

\(=\left(4+\sqrt{5}\right)\left(4-\sqrt{5}\right)\)

=16-5=11

a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)

\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}=-2\)

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

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

\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(A=\sqrt{5}-1-\sqrt{5}-1\)

\(A=-2\)

     \(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(B=\sqrt{5+4\sqrt{5}+4}-\sqrt{5-4\sqrt{5}+4}\)

\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(B=\sqrt{5}+2-\sqrt{5}+2\)

\(B=4\)

Học tốt 

Bài làm:

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

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

\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(A=\sqrt{5}-1-\sqrt{5}-1=-2\)

b) \(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(B=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)

\(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(B=2+\sqrt{5}-\sqrt{5}+2\)

\(B=4\)

Dạ là do bạn rút căn ra ấy ạ:

\(\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

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

\(\Rightarrow\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\left(\sqrt{5}-1\right)-\left(\sqrt{5}+1\right)\)

\(=\sqrt{5}-1-\sqrt{5}-1=-2\)

Chúc bn hc tốt!!!

Lời giải:
a.

\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)

$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$

$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$

b.

$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$

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

$=|\sqrt{3}-3|+|\sqrt{3}+3|$

$=(3-\sqrt{3})+(\sqrt{3}+3)=6$

c.

$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$

$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$

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

\(\dfrac{\sqrt{9-4\sqrt{5}}}{2-\sqrt{5}}=\dfrac{\sqrt{\left(\sqrt{5}-2\right)^2}}{2-\sqrt{5}}=\dfrac{\sqrt{5}-2}{2-\sqrt{5}}=\dfrac{-\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=-1\)

\(\dfrac{\sqrt{9-4\sqrt{5}}}{2-\sqrt{5}}=-1\)

Ta có: \(\sqrt{29+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}\cdot1+1}}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2+2\sqrt{2}+1}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2+2\sqrt{2}\cdot1+1}}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{\left(\sqrt{2}+1\right)^2}}+5\sqrt{2}\)

\(=\sqrt{29+30\left(\sqrt{2}+1\right)}+5\sqrt{2}\)

\(=\sqrt{29+30\sqrt{2}+30}+5\sqrt{2}\)

\(=\sqrt{9+2\cdot3\cdot5\sqrt{2}+50}+5\sqrt{2}\)

\(=\sqrt{\left(3+5\sqrt{2}\right)^2}+5\sqrt{2}\)

\(=3+5\sqrt{2}+5\sqrt{2}=3+10\sqrt{2}\)