Question

tính H = \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)

Answer 1

H=\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)

H=\(\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}\)

H=\(\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2.3.2\sqrt{5}+3^2}}\)

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

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

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

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

H=\(\sqrt{5}-\sqrt{6-2\sqrt{5}}\)

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

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

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

H=\(\sqrt{5}-\left|\sqrt{5}-1\right|\)

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

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

H=1

 

Answer 2

Ta có: \(H=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)

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

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

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

=1