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

Question

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

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2 tháng 2021-10-05T07:31:24+00:00 2 Answers 13 views 0

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