Răspuns:
[tex]\dfrac{1}{6}[/tex]
Explicație pas cu pas:
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[tex]\sqrt{\bigg(1-\dfrac{1}{2}\bigg)\bigg(1-\dfrac{1}{3}\bigg)\bigg(1-\dfrac{1}{4}\bigg)...\bigg(1-\dfrac{1}{36}\bigg)}=\\\\\\\sqrt{\bigg(\dfrac{2}{2}-\dfrac{1}{2}\bigg)\bigg(\dfrac{3}{3}-\dfrac{1}{3}\bigg)\bigg(\dfrac{4}{4}-\dfrac{1}{4}\bigg)...\bigg(\dfrac{36}{36}-\dfrac{1}{36}\bigg)}=\\\\\\\sqrt{\dfrac{1}{\not2}*\dfrac{\not2}{\not3}*\dfrac{\not3}{\not4}*\not..\not.*\dfrac{\not35}{36}}=\\\\\\\sqrt{\dfrac{1}{36}}=\boxed{\boxed{\dfrac{1}{6}}}[/tex]