Tìm tập xác định a, y=căn 3 - x+4/x2 -1 b, y=2 căn 1-x -1/x c,3/căn 4x-x2
2 câu trả lời
\[\begin{array}{l} a)\,\,\,\sqrt {3 - x} + \frac{4}{{{x^2} - 1}}\\ DK:\,\,\,\,\left\{ \begin{array}{l} 3 - x \ge 0\\ {x^2} - 1 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \le 3\\ x \ne \pm 1 \end{array} \right..\\ \Rightarrow D = \left( { - \infty ;\,\,3} \right]\backslash \left\{ { \pm 1} \right\}.\\ b)\,\,\,y = 2\sqrt {1 - x} - \frac{1}{x}\\ DK:\,\,\,\left\{ \begin{array}{l} 1 - x \ge 0\\ x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \le 1\\ x \ne 0 \end{array} \right.\\ \Rightarrow D = \left( { - \infty ;\,\,1} \right]\backslash \left\{ 0 \right\}.\\ c)\,\,\,\frac{3}{{\sqrt {4x - {x^2}} }}\\ DK:\,\,\,4x - {x^2} > 0\\ \Leftrightarrow {x^2} - 4x < 0\\ \Leftrightarrow x\left( {x - 4} \right) < 0\\ \Leftrightarrow 0 < x < 4.\\ \Rightarrow D = \left( {0;\,\,\,4} \right). \end{array}\]
