Question

⇵⊄¬↔←⇵Ф∴←←$\alpha x_{123} \leq \left \{ {{y=2} \atop {x=2}} \right. \sqrt{x} \sqrt{x} \neq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \leq \beta \beta$

Answer 1

Respuesta:2

Explicación: