Đặt \(A=\left(\sqrt{9-\sqrt{17}}\right).\left(\sqrt{9+\sqrt{17}}\right)\)
Ta có: \(A^2=\left[\left(\sqrt{9-\sqrt{17}}\right).\left(\sqrt{9+\sqrt{17}}\right)\right]=\left(9-\sqrt{17}\right).\left(9+\sqrt{17}\right)\)
\(=9^2-\left(\sqrt{17}\right)^2=81-17=64\)
\(=>A=\sqrt{64}=8\)
Xét vế trái:
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\)
\(=\sqrt{\left(\sqrt{\frac{17}{2}}-\sqrt{\frac{1}{2}}\right)^2}.\sqrt{\left(\sqrt{\frac{17}{2}}+\sqrt{\frac{1}{2}}\right)^2}\)
\(=\left|\sqrt{\frac{17}{2}}-\sqrt{\frac{1}{2}}\right|.\left|\sqrt{\frac{17}{2}}+\sqrt{\frac{1}{2}}\right|\)
\(=\left(\sqrt{\frac{17}{2}}-\sqrt{\frac{1}{2}}\right).\left(\sqrt{\frac{17}{2}}+\sqrt{\frac{1}{2}}\right)\)
\(=\frac{17}{2}+\frac{\sqrt{17}}{2}-\frac{\sqrt{17}}{2}-\frac{1}{2}\)
\(=\frac{17}{2}-\frac{1}{2}=8\)
Vậy: \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=8.\)
(Nhớ k cho mình với nha!)
