1/(cosx^3+x) Tim tap XD?

Đáp án:

y=1cosx3+xDK:cosx3+x≠0⇔cosx3≠−x(∗)Vi−1≤cosx3≤1⇒−x∉[−1;1]⇔[−x>1−x<−1⇔[x<−1x>1.VayD=(−∞;−1)∪(1;+∞).y=1cos⁡x3+xDK:cos⁡x3+x≠0⇔cos⁡x3≠−x(∗)Vi−1≤cos⁡x3≤1⇒−x∉[−1;1]⇔[−x>1−x<−1⇔[x<−1x>1.VayD=(−∞;−1)∪(1;+∞).

\[\begin{array}{l} y = \frac{1}{{\cos {x^3} + x}}\\ DK:\,\,\,\cos {x^3} + x \ne 0 \Leftrightarrow \cos {x^3} \ne - x\,\,\,\left( * \right)\\ Vi\,\,\, - 1 \le \cos {x^3} \le 1\\ \Rightarrow - x \notin \left[ { - 1;\,\,1} \right]\\ \Leftrightarrow \left[ \begin{array}{l} - x > 1\\ - x < - 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x < - 1\\ x > 1 \end{array} \right..\\ Vay\,\,\,D = \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right). \end{array}\]