từ đề bài ta có \(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=10\)

=> B=2013. (1+\(\frac{1}{1+2}\) +\(\frac{1}{1+2+3}\) +...+ \(\frac{1}{1+2+3+...+2012}\))

=>B= 2013.(\(\frac{2}{2}\) + \(\frac{2}{2.3}\) +\(\frac{2}{3.4}\) +...+\(\frac{2}{2012.2013}\))

=>B= 2013.2.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +\(\frac{1}{3.4}\) +...+\(\frac{1}{2012.2013}\))

=>B=4026. (1-\(\frac{1}{2}\) +\(\frac{1}{2}\) -\(\frac{1}{3}\) + ...+\(\frac{1}{2012}\) - \(\frac{1}{2013}\))

=>B=4026.(1-\(\frac{1}{2013}\)) 

=>B=4026.\(\frac{2012}{2013}\) => B=2.2012=4024 Vậy B=4024

bn tham khảo link này nha :https://olm.vn/hoi-dap/question/67497.html 

Đặt \(C=B-\frac{1}{2}=\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2012}\)

\(\Rightarrow\frac{3}{2}\cdot C=\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+...+\left(\frac{3}{2}\right)^{2013}\)

\(\Rightarrow\frac{3}{2}C-C=\frac{1}{2}C=\frac{3}{2}+\left(\frac{3}{2}\right)^{2013}\)

\(\Rightarrow C=3+\left(\frac{3}{2}\right)^{2013}\cdot2\)

\(\Rightarrow B=\frac{1}{2}+3+\left(\frac{3}{2}\right)^{2013}\cdot2\)

do đó \(A-B=\left(\frac{3}{2}\right)^{2014}+\frac{7}{2}\)

Đặt phân thức trên là D

=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)

=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)

=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)

=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)

=> D=2015

UwU

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đùa thôi đáp án: 2015 nha bn

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À quên nhớ FOLOW CHO TUI NHA!

d= 2015

Ta có \(A=\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\)

\(\Rightarrow\frac{3}{2}A=\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+....\left(\frac{3}{2}\right)^{2013}\)

\(\Rightarrow\frac{3}{2}A-A=\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}\)hay \(\frac{1}{2}A=\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}\)

Suy ra \(A=2.\text{[}\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}\text{]}\)

Khi đó \(B-A=\frac{\left(\frac{3}{2}\right)^{2013}}{2}-2.\text{[}\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}\text{]}\)

\(A=\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\)

\(\frac{3}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}\)

\(\Rightarrow\frac{3}{2}.A-A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}-\left[\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\right]\)

\(\Rightarrow\frac{1}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}-\frac{3}{2}=\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}\)

\(\Rightarrow A=2.\left(\frac{3}{2}\right)^{2013}-\frac{5}{2}\)

\(B-A=\frac{1}{2}.\left(\frac{3}{2}\right)^{2013}-2.\left(\frac{3}{2}\right)^{2013}+\frac{5}{2}=-\left(\frac{3}{2}\right)^{2014}+\frac{5}{2}\)