Bài 4:
\(A=\sqrt{6+2\sqrt3+2\sqrt2+2\sqrt6}\)
\(=\sqrt{3+2+1+2\cdot\sqrt3\cdot1+2\cdot\sqrt2\cdot1+2\cdot\sqrt3\cdot\sqrt2}\)
\(=\sqrt{\left(\sqrt3+\sqrt2+1\right)^2}=\sqrt3+\sqrt2+1\)
Bài 2:
a: \(\sqrt{8-2\sqrt7}=\sqrt{\left(\sqrt7-1\right)^2}=\sqrt7-1\)
b: \(\sqrt{4-\sqrt7}-\sqrt{4+\sqrt7}\)
\(=\frac{1}{\sqrt2}\left(\sqrt{8-2\sqrt7}-\sqrt{8+2\sqrt7}\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt7-1\right)^2}-\sqrt{\left(\sqrt7+1\right)^2}\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt7-1-\sqrt7-1\right)=-\frac{2}{\sqrt2}=-\sqrt2\)
c: \(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)
\(=\frac{1}{\sqrt2}\left(\sqrt{6-2\sqrt5}+\sqrt{6+2\sqrt5}\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt5-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}\right)=\frac{1}{\sqrt2}\left(\sqrt5-1+\sqrt5+1\right)=\frac{1}{\sqrt2}\cdot2\sqrt5=\sqrt{10}\)
VD3: \(A=\sqrt{2-\sqrt3}+\sqrt{2+\sqrt3}\)
\(=\frac{1}{\sqrt2}\left(\sqrt{4-2\sqrt3}+\sqrt{4+2\sqrt3}\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3-1\right)^2}+\sqrt{\left(\sqrt3+1\right)^2}\right)\)
\(=\frac{1}{\sqrt2}\left(\sqrt3-1+\sqrt3+1\right)=\frac{2\sqrt3}{\sqrt2}=\sqrt6\)
