\(\sqrt{8-\sqrt{15}}-\sqrt{8+\sqrt{15}}=\dfrac{\sqrt{16-2\sqrt{15}}-\sqrt{16+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{15}-1\right)^2}-\sqrt{\left(\sqrt{15}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{15}-1-\sqrt{15}-1}{\sqrt{2}}=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)
Đặt \(A=\sqrt{8-\sqrt{15}}-\sqrt{8+\sqrt{15}}\)
\(\Rightarrow A^2=\left(\sqrt{8-\sqrt{15}}-\sqrt{8+\sqrt{15}}\right)^2\)
\(\Leftrightarrow A^2=8-\sqrt{15}-2\sqrt{\left(8-\sqrt{15}\right)\left(8+\sqrt{15}\right)}+8+\sqrt{15}\)
\(\Leftrightarrow A^2=16-2\sqrt{8^2-\left(\sqrt{15}\right)^2}\)
\(\Leftrightarrow A^2=16-2\sqrt{49}=16-14=2\)
Vì \(\left(8-\sqrt{15}\right)< \left(8+\sqrt{15}\right)\Leftrightarrow\sqrt{8-\sqrt{15}}< 8+\sqrt{15}\)
suy ra \(\sqrt{8-\sqrt{15}}-\sqrt{8+\sqrt{15}}< 0\)Vậy A<0
Ta có\(A^2=2\) và A<0
suy ra A=\(-\sqrt{2}\)
