Question

Cho a = \(\sqrt{4+\sqrt{10+2\sqrt{5}}}\) và b = \(\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

Tính \(a^2+b^2\)và ab từ đó suy ra a + b

Answer 1

\(a^2+b^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}=8\)

\(ab=\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}=\sqrt{16-\left(10+2\sqrt{5}\right)}\)

\(=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

\(\Rightarrow\left(a+b\right)^2=a^2+b^2+2ab=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)

\(\Rightarrow a+b=\sqrt{5}+1\) (do \(a;b>0\Rightarrow a+b>0\))