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Dang thi my dung
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Bùi Minh Hằng
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Phương Linh Bùi
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☆ĐP◈Replay-Music
9 tháng 5 2019 lúc 20:38

\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)

\(\frac{B}{A}=\frac{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)

\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)

\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)

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Huỳnh Thị Thu Uyên
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soyeon_Tiểubàng giải
14 tháng 9 2016 lúc 12:09

\(A=\left(1+2\right).\frac{1}{2}+\left(1+2+3\right).\frac{1}{3}+...+\left(1+2+3+...+2016\right).\frac{1}{2016}\)

\(A=\left(1+2\right).2:2.\frac{1}{2}+\left(1+3\right).3:2.\frac{1}{3}+...+\left(1+2016\right).2016:2.\frac{1}{2016}\)

\(A=3:2+4:2+...+2017:2\)

\(A=3.\frac{1}{2}+4.\frac{1}{2}+...+2017.\frac{1}{2}\)

\(A=\frac{1}{2}.\left(3+4+...+2017\right)\)

\(A=\frac{1}{2}.\left(3+2017\right).2015:2\)

\(A=\frac{1}{2}.2020.2015.\frac{1}{2}\)

\(A=505.2015=1017575\)

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Nguyễnn Vũtháibìnhh
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Nguyễn Lê Phước Thịnh
22 tháng 3 2021 lúc 13:42

Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)

\(=2017\)

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nguyễn xuân trường
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Tienanh nguyễn
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Đặng Hoàng Anh Khoa
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Lê Anh Quân
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Nguyễn Ngọc Kim Ngân
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