a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)

b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)

c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)

d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)

e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)

f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)

Cho mình sửa đề xí ạ! 

b) \(\frac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\)

\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)

a, \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{17-2.3.2\sqrt{2}}-\sqrt{17+2.3.2\sqrt{2}}\)

\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)

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

\(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)

b, \(\sqrt{31-12\sqrt{3}}-\sqrt{31+12\sqrt{3}}\)

\(=\sqrt{31-2.2.3\sqrt{3}}-\sqrt{31+2.2.3\sqrt{3}}\)

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

\(=3\sqrt{3}-2-3\sqrt{3}-2=-4\)

1.\(\sqrt{11-4\sqrt{7}}=\sqrt{11-2\cdot2\sqrt{7}}\)=\(\sqrt{7-2\cdot2\cdot\sqrt{7}+4}=\sqrt{\left(\sqrt{7}-2\right)^2}\)=\(\sqrt{7}-2\)

2.\(\left(3-\sqrt{9}\right)\sqrt{11+6\sqrt{6}}=\left(3-3\right)\sqrt{11+6\sqrt{6}}\)=\(0\cdot\sqrt{11+6\sqrt{6}}=0\)

3.\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\)\(\sqrt{15-2\cdot3\cdot\sqrt{6}}+\sqrt{35-2\cdot2\cdot3\cdot\sqrt{2}\cdot\sqrt{3}}\)

                                                      =\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{35-2\cdot\left(2\sqrt{2}\right)\left(3\sqrt{3}\right)}\)=\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{27-2\cdot\left(2\sqrt{2}\right)\left(3\sqrt{3}\right)+8}\)

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

đoạn cuối thiếu dấu"+"

\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)

\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)

\(A=\sqrt{36}-\sqrt{4}=6-2=4\)

\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+...+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}+\sqrt{36}}\)

\(=-\sqrt{4}+\sqrt{5}-\sqrt{5}+\sqrt{6}-...-\sqrt{35}+\sqrt{36}\)

\(=6-2=4\)

1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)

 

\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)

\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)

\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)

\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)

\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)

\(=\)\(3\sqrt{6}+9\)

Chúc bạn học tốt ~ 

\(b)\)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)

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

\(=\)\(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\) ( vì \(2=\sqrt{4}>\sqrt{3}\) ) 

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

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

\(=\)\(1\)

Chúc bạn học tốt ~ 

PS : mới lớp 8 sai thì thông cảm >.< 

\(\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}\) + \(\sqrt{\left(\sqrt{24}-\sqrt{9}\right)^2}\)

\(\sqrt{9}-\sqrt{6}\) + \(\sqrt{24}-\sqrt{9}\) vì  \(\sqrt{9}>\sqrt{6}\),\(\sqrt{24}>\sqrt{9}\)

\(3-\sqrt{6}\) + \(2\sqrt{6}-3\)

= \(\sqrt{6}\)