Answer:
hbhuobhul
Explanation:
hg k[tex]x^{2}\sqrt{x} \sqrt[n]{x} \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \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] \neq x^{2} \sqrt{x} \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \frac{x}{y} \pi[/tex]h hk h⇄⇆⇒Ф∫βπФ∈∉⊥∡¬←Δ⇒×ω⇆⊕⊕║⇄ωФ⇄⇄㏑
