Question

???????? как решить​

Answer 1

$(1-\dfrac{1}{n^2})=\dfrac{n^2-1}{n^2}=\dfrac{(n-1)(n+1)}{n^2}\\ (1-\dfrac{1}{5^2})(1-\dfrac{1}{6^2})...(1-\dfrac{1}{14^2})=\dfrac{4*6}{5^2}*\dfrac{5*7}{6^2}...\dfrac{13*15}{14^2}=4*5*14*15*\dfrac{6^2*7^2*...*13^2}{5^2*6^2*...*14^2}=\dfrac{4*5*14*15}{5^2*14^2}=\dfrac{4*15}{5*14}=\dfrac{6}{7}\\ \dfrac{6}{7}(x-1)=\dfrac{3}{7}=>x-1=\dfrac{1}{2}=>x=1,5$

Ответ: А) 1,5