Tìm tập xác định D của hàm số y = x/ x- căn x -6 y= căn tất cả của căn x ² +2x+2 -(x+1)

\[\begin{array}{l} a)\,\,\,y = \frac{x}{{x - \sqrt x - 6}}\\ DK:\,\,\,\left\{ \begin{array}{l} x \ge 0\\ x - \sqrt x - 6 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ \left( {x + 2} \right)\left( {x - 3} \right) \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne - 2\\ x \ne 3 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne 3 \end{array} \right.\\ \Rightarrow D = \left[ {0; + \infty } \right)\backslash \left\{ 3 \right\}.\\ b)\,\,y = \sqrt {\sqrt {{x^2} + 2x + 2} - \left( {x + 1} \right)} \\ DK:\,\,\,\sqrt {{x^2} + 2x + 2} - \left( {x + 1} \right) \ge 0\\ \Leftrightarrow \sqrt {{x^2} + 2x + 2} \ge x + 1\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 \le 0\\ \left\{ \begin{array}{l} x + 1 \ge 0\\ {x^2} + 2x + 2 \ge {\left( {x + 1} \right)^2} \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x \le - 1\\ \left\{ \begin{array}{l} x \ge - 1\\ {x^2} + 2x + 2 \ge {x^2} + 2x + 1 \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x \le - 1\\ \left\{ \begin{array}{l} x \ge - 1\\ 2 > 1\,\,\,\forall x \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x \le - 1\\ x \ge - 1 \end{array} \right. \Rightarrow x \in R.\\ \Rightarrow D = R. \end{array}\]