Bài 1 :
a, ĐKXĐ : \(\left\{{}\begin{matrix}x-\sqrt{2x+1}\ne0\\2x+1\ge0\\x-\sqrt{2x+1}\ge0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}2x+1\ge0\\x>\sqrt{2x+1}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ge-\frac{1}{2}\\x>\sqrt{2x+1}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ge-\frac{1}{2}\\x>\sqrt{2x+1}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ge-\frac{1}{2}\\x^2-2x-1>0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ge-\frac{1}{2}\\\left[{}\begin{matrix}x>1+\sqrt{2}\\x< 1-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
=> \(x>1+\sqrt{2}\)
b, ĐKXĐ : \(1-16x^2\ge0\)
=> \(x^2\le\frac{1}{16}\)
=> \(-\frac{1}{4}\le x\le\frac{1}{4}\)
Bài 2 :
a, Ta có : \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{2^2+2.2\sqrt{3}+\left(\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{5^2-2.5\sqrt{3}+\left(\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}=\sqrt{25}=5\)
em cảm mơn nhìu nhìu 
