Question

Tìm TXD

a) y= sin\(\sqrt{x^2-2x}\)

b) y= \(\dfrac{2sinx}{cos2x-1}\)

Answer 1

\(a,y\)\(=sin\sqrt{x^2-2x}\)

Đkxđ: \(\sqrt{x^2-2x}\in R\)

\(\Leftrightarrow x^2-2x\ge0\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x\le0\\x\ge2\end{matrix}\right.\)

\(\Rightarrow D:(-\infty;0]\cup[2;+\infty)\)

\(b,y\)\(=\dfrac{2sinx}{cos2x-1}\)

Đkxđ: cos2x-1\(\in R\)

\(\Leftrightarrow cos2x-1\ne0\)

\(\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

\(\Rightarrow D:R\backslash\left\{\dfrac{k\pi}{2};k\in\right\}\)