Răspuns:
Explicație pas cu pas:
a²=2; b⁶=11. Deci [tex]a=\sqrt{2}; ~~b=\sqrt[6]{11}.~~( \sqrt{2})^{6}<(\sqrt[6]{11})^{6},~~2^{3}<11 ~adevarat.\\ Deci~a<b;~rezulta~m=\sqrt{2},~iar ~M=\sqrt[6]{11}.\\m<\sqrt[3]{x}<M,~~ \sqrt{2}<\sqrt[3]{x}<\sqrt[6]{11},~|^3,~~\sqrt{2^{3}}<(\sqrt[3]{x})^{3}<\sqrt[6]{11^{3}},~\sqrt{8}<x<\sqrt{11}[/tex]
Deoarece x∈Z, ⇒x=3. Deci H={3}.