Tìm tỉ số \(\frac{A}{B}\)biết
a,\(A=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)
\(B=\frac{7}{19.31}+\frac{5}{19.43}+\frac{3}{23.43}+\frac{11}{23.57}\)
b,\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\)
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
GIẢI RÕ NHA ! NHANH NHANH MÌNH CẦN GẤP !+_+
b) Ta có:
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(\Rightarrow B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(\Rightarrow B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(\Rightarrow B=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}=\frac{1}{2017}\)
Vậy \(\frac{A}{B}=\frac{1}{2017}\)