thương A chia B là \(\frac{A}{B}\)
ta có :
\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)
\(=\frac{10-1}{1}+\frac{10-2}{2}+\frac{10-3}{3}+...+\frac{10-9}{9}\)
\(=10+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{9}{9}\right)\)
\(=10+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}-9\)
\(=\left(10-9\right)+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)
\(=1+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)
\(=10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)=10B\)
=> A : B = 10
Vậy A : B = 10
ta có: \(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}.\)
\(A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)\)
\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)
\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}=10\)
