Đáp án:
$\begin{array}{l}
1)Dkxd:\left\{ \begin{array}{l}
{x^2} - 1 \ge 0\\
x - 1 \ge 0
\end{array} \right. \Rightarrow x \ge 1\\
\sqrt {{x^2} - 1} - \sqrt {x - 1} = 0\\
\Rightarrow \sqrt {{x^2} - 1} = \sqrt {x - 1} \\
\Rightarrow {x^2} - 1 = x - 1\\
\Rightarrow {x^2} - x = 0\\
\Rightarrow x\left( {x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {ktm} \right)\\
x = 1\left( {tm} \right)
\end{array} \right.\\
Vay\,x = 1\\
2)a){4^2} = 16\\
{\left( {2\sqrt 3 } \right)^2} = 12\\
\Rightarrow 4 > 2\sqrt 3 \\
b){\left( { - \sqrt 5 } \right)^2} = 5\\
{\left( { - 2} \right)^2} = 4\\
\Rightarrow - \sqrt 5 < - 2\\
3)a)\left( {2 - \sqrt 3 } \right).\left( {2 + \sqrt 3 } \right)\\
= {2^2} - {\left( {\sqrt 3 } \right)^2}\\
= 4 - 3 = 1\\
b)\left( {\sqrt {2006} - \sqrt {2005} } \right).\left( {\sqrt {2006} + \sqrt {2005} } \right)\\
= 2006 - 2005 = 1\\
\Rightarrow \sqrt {2006} - \sqrt {2005} = \dfrac{1}{{\sqrt {2006} + \sqrt {2005} }}\\
4)\sqrt {25 + 9} = \sqrt {33} \\
\sqrt {25} + \sqrt 9 = 5 + 3 = 8 = \sqrt {64} \\
\Rightarrow \sqrt {25 + 9} < \sqrt {25} + \sqrt 9 \\
5)\\
{\left( {\sqrt {a + b} } \right)^2} = a + b\\
{\left( {\sqrt a + \sqrt b } \right)^2} = a + 2\sqrt {ab} + b = a + b + 2\sqrt {ab} \\
Do:a,b > 0\\
\Rightarrow 2\sqrt {ab} > 0\\
\Rightarrow a + b + 2\sqrt {ab} > a + b\\
\Rightarrow {\left( {\sqrt {a + b} } \right)^2} < {\left( {\sqrt a + \sqrt b } \right)^2}\\
\Rightarrow \sqrt {a + b} < \sqrt a + \sqrt b
\end{array}$
