Question

va rog ajutati-ma! chiar nu stiu.

Answer 1

Răspuns:

12

Explicație pas cu pas:

$\sqrt{\Big(1 + \dfrac{1}{2}\Big)\Big(1 + \dfrac{1}{3}\Big)\Big(1 + \dfrac{1}{4}\Big)\Big(1 + \dfrac{1}{5}\Big)...\Big(1 + \dfrac{1}{285}\Big)\Big(1 + \dfrac{1}{286}\Big)\Big(1 + \dfrac{1}{287}\Big)} =$

$= \sqrt{\dfrac{ \not3}{2} \cdot \dfrac{ \not4}{ \not3} \cdot \dfrac{ \not5}{ \not4} \cdot \dfrac{ \not6}{ \not5} \cdot ... \cdot \dfrac{286}{285} \cdot \dfrac{287}{286} \cdot \dfrac{288}{287}}$

$= \sqrt{\dfrac{288}{2}} = \sqrt{144} = \bf 12$