Tính: \(\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\) 14/08/2021 Bởi Gabriella Tính: \(\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
Ta có: \(\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\) \(=\sqrt{1+\sqrt{3+\sqrt{12+2\cdot2\sqrt{3}\cdot1+1}}}+\sqrt{1-\sqrt{3-\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\) \(=\sqrt{1+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}}+\sqrt{1-\sqrt{3-\sqrt{\left(2\sqrt{3}-1\right)^2}}}\) \(=\sqrt{1+\sqrt{3+\left|2\sqrt{3}+1\right|}}+\sqrt{1-\sqrt{3-\left|2\sqrt{3}-1\right|}}\) \(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)(Vì \(2\sqrt{3}>1>0\)) \(=\sqrt{1+\sqrt{4+2\sqrt{3}}}+\sqrt{1-\sqrt{3-2\sqrt{3}+1}}\) \(=\sqrt{1+\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}+\sqrt{1-\sqrt{\left(\sqrt{3}-1\right)^2}}\) \(=\sqrt{1+\sqrt{\left(\sqrt{3}+1\right)^2}}+\sqrt{1-\sqrt{\left(\sqrt{3}-1\right)^2}}\) \(=\sqrt{1+\left|\sqrt{3}+1\right|}+\sqrt{1-\left|\sqrt{3}-1\right|}\) \(=\sqrt{1+\sqrt{3}+1}+\sqrt{1-\left(\sqrt{3}-1\right)}\)(Vì \(\sqrt{3}>1>0\)) \(=\sqrt{2+\sqrt{3}}+\sqrt{1-\sqrt{3}+1}\) \(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\) \(=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\) \(=\frac{\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}}\) \(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\) \(=\frac{\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|}{\sqrt{2}}\) \(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}\)(Vì \(\sqrt{3}>1>0\)) \(=\frac{2\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{12}}{\sqrt{2}}=\sqrt{6}\) Bình luận
Xem hình…
Ta có: \(\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
\(=\sqrt{1+\sqrt{3+\sqrt{12+2\cdot2\sqrt{3}\cdot1+1}}}+\sqrt{1-\sqrt{3-\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)
\(=\sqrt{1+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}}+\sqrt{1-\sqrt{3-\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(=\sqrt{1+\sqrt{3+\left|2\sqrt{3}+1\right|}}+\sqrt{1-\sqrt{3-\left|2\sqrt{3}-1\right|}}\)
\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)(Vì \(2\sqrt{3}>1>0\))
\(=\sqrt{1+\sqrt{4+2\sqrt{3}}}+\sqrt{1-\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{1+\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}+\sqrt{1-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{1+\sqrt{\left(\sqrt{3}+1\right)^2}}+\sqrt{1-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{1+\left|\sqrt{3}+1\right|}+\sqrt{1-\left|\sqrt{3}-1\right|}\)
\(=\sqrt{1+\sqrt{3}+1}+\sqrt{1-\left(\sqrt{3}-1\right)}\)(Vì \(\sqrt{3}>1>0\))
\(=\sqrt{2+\sqrt{3}}+\sqrt{1-\sqrt{3}+1}\)
\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}\)(Vì \(\sqrt{3}>1>0\))
\(=\frac{2\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{12}}{\sqrt{2}}=\sqrt{6}\)