\[\begin{array}{l} y = \frac{2}{{\sqrt {{{\sin }^2}x - 2\sin x + m - 1} }}\\ Hs\,\,xd\,\,tren\,\,\,R \Leftrightarrow {\sin ^2}x - 2\sin x + m - 1 > 0\,\,\forall x \in R\,\,\,\left( * \right)\\ Dat\,\,\sin x = t\\ Ta\,\,co:\,\,\, - 1 \le \sin x \le 1 \Rightarrow - 1 \le t \le 1.\\ \Rightarrow \left( * \right) \Leftrightarrow {t^2} - 2t + m - 1 > 0\,\,\forall t \in \left[ { - 1;\,\,1} \right].\\ \Leftrightarrow {t^2} - 2t + 1 + m - 2 > 0\,\,\,\forall t \in \left[ { - 1;\,1} \right]\\ \Leftrightarrow {\left( {t - 1} \right)^2} > 2 - m\,\,\forall t \in \left[ { - 1;\,\,1} \right]\\ \Leftrightarrow 2 - m < \mathop {Min}\limits_{\left[ { - 1;\,\,1} \right]} {\left( {t - 1} \right)^2}\\ Voi\,\,\,t \in \left[ { - 1;\,\,1} \right] \Rightarrow 0 \le {\left( {t - 1} \right)^2} \le 4\\ \Rightarrow 2 - m < 0\\ \Leftrightarrow m > 2.\\ Vay\,\,m > 2\,\,thi\,\,hs\,\,xd\,\,tren\,\,R. \end{array}\]

Đáp án:

Giải thích các bước giải:

$$\eqalign{ & y = \,{2 \over {\sqrt {\sin 2x - 2\sin x + m - 1} }}\,\,xac\,dinh/R \cr & \Leftrightarrow {\sin ^2}x - 2\sin x + m - 1 > 0\,\,\forall x \in R \cr & Dat\,\,t = \sin x \Rightarrow t \in \left[ { - 1;1} \right] \cr & \Rightarrow {t^2} - 2t + m - 1 > 0\,\,\forall t \in \left[ { - 1;1} \right] \cr & Tu\,\,BBT \Rightarrow - 2 + m > 0 \Leftrightarrow m > 2 \cr} $$