Question

можно немецкий м.м Сидоренко стр.58 упр 3

Answer 1

$\left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \sqrt{x} \leq \neq \alpha \geq \pi$