Question

Chứng minh đẳng thức sau với \(b\ge0;a\ge\sqrt{b}\)

 \(\sqrt{a+\sqrt{b}}\mp\sqrt{a-\sqrt{b}}=\sqrt{2\left(a\mp\sqrt{a^2-b}\right)}\)

Answer 1

\(\Rightarrow\left(\sqrt{a+\sqrt{b}}\mp\sqrt{a-\sqrt{b}}\right)^2=\left(\sqrt{2\left(a\mp\sqrt{a^2-b}\right)}\right)^2\Leftrightarrow a+\sqrt{b}+a-\sqrt{b}\mp2\sqrt{\left(a+\sqrt{b}\right)\cdot\left(a-\sqrt{b}\right)}=2a\mp2\sqrt{a^2-b}\Leftrightarrow2a\mp2\sqrt{a^2-b}=2a\mp2\sqrt{a^2-b}\) (luôn đúng) \(\Rightarrowđpcm\)