Question

Cho a, b là hai số dương thoả mãn \(a^2-b^2\ge0\). Chứng minh:

\(\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}+\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}=\sqrt{a+\sqrt{b}}\)

Answer 1

\(\left(\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}+\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}\right)^2\\ =\dfrac{a+\sqrt{a^2-b}+a-\sqrt{a^2-b}}{2}+2\sqrt{\dfrac{\left(a+\sqrt{a^2-b}\right)\left(a-\sqrt{a^2-b}\right)}{4}}\\ =\dfrac{2a}{2}+2\sqrt{\dfrac{a^2-a^2+b}{4}}\\ =a+2\sqrt{\dfrac{b}{4}}=a+\dfrac{2\sqrt{b}}{2}=a+\sqrt{b}\\ \Rightarrow\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}+\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}=\sqrt{a+\sqrt{b}}\)