b)
$\begin{array}{l} DKXD:\left\{ \begin{array}{l} \frac{1}{{{x^2} - 2x + 1}} \ge 0\\ {x^2} - 2x + 1 \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \frac{1}{{{{\left( {x - 1} \right)}^2}}} \ge 0(luôn \,đúng)\\ {\left( {x - 1} \right)^2} \ne 0 \end{array} \right. \Leftrightarrow x \ne 1 \end{array}$
2.1
$\begin{array}{l} ab + b\sqrt a + \sqrt a + 1\\ = b\left( {\sqrt a + a} \right) + \left( {\sqrt a + 1} \right)\\ = b\sqrt a \left( {\sqrt a + 1} \right) + \left( {\sqrt a + 1} \right)\\ = \left( {\sqrt a + 1} \right)\left( {1 + b\sqrt a } \right)\\ b)4a + 1\left( {a < 0} \right)\\ = 1 - \left( { - 4a} \right) = \left( {1 - \sqrt { - 2a} } \right)\left( {1 + \sqrt { - 2a} } \right)\\ \end{array}$
2.2 Điều kiện xác định $x\ge -1$
$\begin{array}{l} \sqrt {9x + 9} + \sqrt {x + 1} = 20\\ \Leftrightarrow \sqrt {9\left( {x + 1} \right)} + \sqrt {x + 1} = 20\\ \Leftrightarrow 3\sqrt {x + 1} + \sqrt {x + 1} = 20\\ \Leftrightarrow 4\sqrt {x + 1} = 20\\ \Leftrightarrow \sqrt {x + 1} = 5\\ \Leftrightarrow x + 1 = 25 \Leftrightarrow x = 24\\ \Rightarrow S = \left\{ {24} \right\} \end{array}$
