X
xuan_nam


Chứng minh đẳng thức sau với $b \ge 0; a \ge \sqrt{b}$
a) $\sqrt{a + \sqrt{b}} \pm \sqrt{a - \sqrt{b}} = \sqrt{2(a \pm \sqrt{a^2 - b})}$
b) $\sqrt{a \pm \sqrt{b}} = \sqrt{\dfrac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\dfrac{a - \sqrt{a^2 - b}}{2}}$
a) $\sqrt{a + \sqrt{b}} \pm \sqrt{a - \sqrt{b}} = \sqrt{2(a \pm \sqrt{a^2 - b})}$
b) $\sqrt{a \pm \sqrt{b}} = \sqrt{\dfrac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\dfrac{a - \sqrt{a^2 - b}}{2}}$