Bài 1:
\(A=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=B+C\)
\(B=\sqrt{\frac{\left(a+\sqrt{b}\right)+2\sqrt{\left(a-\sqrt{b}\right)\left(a+\sqrt{b}\right)}+\left(a-\sqrt{b}\right)}{4}}=\frac{1}{2}.\sqrt{\left[\sqrt{\left(a+\sqrt{b}\right)}+\sqrt{\left(a-\sqrt{b}\right)}\right]^2}\)
\(B=\frac{1}{2}\left[\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}\right]\)(1)
\(C=\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=\frac{1}{2}.!\left[\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}\right]!\) do \(a\ge\sqrt{b}\ge0\) \(\Rightarrow C=\frac{1}{2}\left[\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}\right]\)(2)
(1) cộng (2)=> dpcm
2). Đặt \(\sqrt{x^2+x+1}=t;t\ge0\Rightarrow t^2=x^2+x+1\)
\(\Leftrightarrow x^2+x+1-3\sqrt{....}+2=0\Leftrightarrow t^2-3t+2=0\)
{a+b+c=0}
\(\left\{\begin{matrix}t=1\Leftrightarrow x^2+x+1=1\left(1\right)\\t=2\Leftrightarrow x^2+x+1=4\left(2\right)\end{matrix}\right.\)
(1) \(\Rightarrow\left[\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
(2) \(\Delta=1+4.3=13\Rightarrow\) \(\left\{\begin{matrix}x=\frac{-1-\sqrt{13}}{2}\\x=\frac{-1+\sqrt{13}}{2}\end{matrix}\right.\)
\(\left|\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}\right|\) ok !