\( A = \sqrt {\dfrac{{8 + \sqrt {15} }}{2}} + \sqrt {\dfrac{{8 - \sqrt {15} }}{2}} \\ \Rightarrow {A^2} = \dfrac{{8 + \sqrt {15} }}{2} + 2\sqrt {\dfrac{{8 + \sqrt {15} }}{2}.\dfrac{{8 - \sqrt {15} }}{2}} + \dfrac{{8 - \sqrt {15} }}{2}\\ = 8 + 2\sqrt {\dfrac{{\left( {8 + \sqrt {15} } \right)\left( {8 - \sqrt {15} } \right)}}{4}} = 8 + 7 = 15 \Rightarrow A = \sqrt {15} \)
Ta có: \(A^2=\frac{8+\sqrt{15}}{2}+\frac{8-\sqrt{15}}{2}+2\sqrt{\frac{8+\sqrt{15}}{2}.\frac{8-\sqrt{15}}{2}}\)
\(A^2=8+2\sqrt{\frac{64-15}{4}}\)
\(A^2=8+2\sqrt{\frac{49}{4}}\)
\(A^2=8+7=15\)
Mà A > 0 nên \(A=\sqrt{15}\)
- Ta có : \(A=\sqrt{\frac{8+\sqrt{15}}{2}}+\sqrt{\frac{8-\sqrt{15}}{2}}\)
=> \(A=\frac{\sqrt{2\left(8+\sqrt{15}\right)}}{2}+\frac{\sqrt{2\left(8-\sqrt{15}\right)}}{2}\)
=> \(A=\frac{\sqrt{15+2\sqrt{15}+1}}{2}+\frac{\sqrt{15-2\sqrt{15}+1}}{2}\)
=> \(A=\frac{\sqrt{\left(\sqrt{15}+1\right)^2}}{2}+\frac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2}\)
=> \(A=\frac{\sqrt{15}+1+\sqrt{15}-1}{2}=\frac{2\sqrt{15}}{2}=\sqrt{15}\)
Lời giải:
\(A=\sqrt{\frac{16+2\sqrt{15}}{4}}+\sqrt{\frac{16-2\sqrt{15}}{4}}=\sqrt{\frac{15+1+2\sqrt{15}}{4}}+\sqrt{\frac{15+1-2\sqrt{15}}{4}}\)
\(=\sqrt{\frac{(\sqrt{15}+1)^2}{2^2}}+\sqrt{\frac{(\sqrt{15}-1)^2}{2^2}}=\frac{\sqrt{15}+1}{2}+\frac{\sqrt{15}-1}{2}=\sqrt{15}\)
