Đặt : \(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
=> \(A^2=16-2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(=16-2\sqrt{8^2-4\left(10+2\sqrt{5}\right)}\)
\(=16-2\sqrt{24-8\sqrt{5}}\)
\(=16-2\sqrt{20-2.2\sqrt{5}.2+4}\)
\(=16-2\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(=16-2\left(2\sqrt{5}-2\right)=20-4\sqrt{5}\)
=> \(A=\sqrt{20-4\sqrt{5}}\)
\(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\)+\(\dfrac{8}{1-\sqrt{5}}\)
\(\dfrac{5+\sqrt{7}}{9-\sqrt{23+8\sqrt{7}}}\)+\(\dfrac{5-\sqrt{7}}{2+\sqrt{16+6\sqrt{7}}}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\)+\(\dfrac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}\)
Chứng minh :
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
a,\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}=\dfrac{8}{1-\sqrt{5}}\)
b,\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
Chứng minh rằng
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
Chứng minh \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}=\sqrt{2}+\sqrt{10}\)
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right).\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(8+2\sqrt{16-10-2\sqrt{5}}\)
\(8+2\sqrt{6-2\sqrt{5}}\)
\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(8+2\left(\sqrt{5}-1\right)\)
\(8+2\sqrt{5}-2\)
\(6+2\sqrt{5}\)
\(\left(\sqrt{5+1}\right)^2\)
\(\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(|\sqrt{5}+1|\)
\(\sqrt{5}+1\)
\(\text{thư ngu như chó}\)
\(\sqrt{8+2\sqrt{10+2\sqrt{5}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}}\)
Tính: \(C=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}-4\sqrt{3+\sqrt{5}}\)
