y = căn 1 + x / 1 - x y =sin 2x/ x - 1 y = căn 1 - cos ^2 x tìm tập xác định

Đáp án:

Giải thích các bước giải:

$$\eqalign{ & y = \sqrt {{{1 + x} \over {1 - x}}} \cr & DKXD:\,\,\,\left\{ \matrix{ {{1 + x} \over {1 - x}} \ge 0 \hfill \cr 1 - x \ne 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ - 1 \le x < 1 \hfill \cr x \ne 1 \hfill \cr} \right. \Leftrightarrow - 1 \le x < 1 \cr & \Rightarrow D = \left[ { - 1;1} \right) \cr & \cr & y = {{\sin 2x} \over {x - 1}} \cr & DKXD:\,\,x - 1 \ne 0 \Leftrightarrow x \ne 1 \cr & \Rightarrow D = R\backslash \left\{ 1 \right\} \cr & \cr & y = \sqrt {1 - {{\cos }^2}2x} \cr & DKXD:\,\,1 - {\cos ^2}2x \ge 0 \Leftrightarrow {\cos ^2}2x \le 1\,\,\left( {luon\,\,dung} \right) \cr & \Rightarrow D = R \cr}$$

$\begin{array}{l} a)\,\,\,y = \sqrt {\frac{{1 + x}}{{1 - x}}} \\ DKXD:\,\,\,\frac{{1 + x}}{{1 - x}} \ge 0\\ \Leftrightarrow \frac{{x + 1}}{{x - 1}} \le 0 \Leftrightarrow - 1 \le x < 1.\\ \Rightarrow D = \left[ { - 1;\,\,1} \right).\\ b)\,\,\,y = \frac{{\sin 2x}}{{x - 1}}\\ DKXD:\,\,x - 1 \ne 0 \Leftrightarrow x \ne 1.\\ \Rightarrow D = R\backslash \left\{ 1 \right\}.\\ c)\,\,y = \sqrt {1 - {{\cos }^2}x} = \sqrt {{{\sin }^2}x} .\\ D = R. \end{array}$