Question

$\sqrt{x} x^{2} x^{2} \int\limits^a_b {x} \, dx \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$

Answer 1

Explanation:

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Answer 2

Answer:

$\pi \neq \alpha \geq \geq \sqrt{x} \sqrt[n]{x} \sqrt[n]{x} \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \geq \leq \neq \sqrt{x} \sqrt{x}$

Explanation: