\(\sqrt{9-6\sqrt{6}+6}+\sqrt{9+6\sqrt{6}+6}\\ =\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}=6\)

sửa lại lúc  nhìm nhầm 

\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{24-12\sqrt{6}+3}\\ =3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{24-12\sqrt{6}+3}\\ =3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

Đây là sửa lại ko phải chứng minh đâu nhé

\(=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{6^2}}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}\)

\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4+\sqrt{15}}\)

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

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

 \(VT\Leftrightarrow\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2=VP\left(dpcm\right)\)

\(\left(\dfrac{\sqrt{30}-\sqrt{20}}{\sqrt{3}-\sqrt{2}}-\dfrac{6}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)

\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}\)

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

\(=5-3=2\)

Lời giải:

$(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}$

$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}$

$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}$

$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})$

$=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})$

$=2(4^2-15)=2$ (đpcm)

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

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

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

a)\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=1\)\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=1\)

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

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

\(\Leftrightarrow\sqrt{1}=1\) (đpcm)

\(\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}-1=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{3}}-1=0\)

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

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

\(\Leftrightarrow3-2-1=0\) (đpcm)

\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}=\sqrt{3+2+1+\sqrt{2^2.2.3}+\sqrt{2^2.3}+\sqrt{2^2.2}}\)

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

(áp dụng hằng đẳng thức (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc)

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

\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)

\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)

\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)

\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)

\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)

\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)