\(\frac{999}{1000}+\frac{998}{1000}+......+\frac{1}{1000}\)

\(=\frac{999+998+997+........+1}{1000}\)

\(=\frac{499500}{1000}=\frac{999}{2}\)

\(\frac{999}{1000}+\frac{998}{1000}+.....+\frac{1}{1000}\)

= \(\frac{999+998+.....+1}{1000}\)(cách tính phép tính này rất đơn giản,chỉ việc lấy(999 + 1) x 999 : 2 = ?)

= \(\frac{499500}{1000}=\frac{999}{2}\)

k mình đi! mình trả lời rõ ràng nhất mà

1/1000 + ... + 997/1000 + 998/1000 + 999/1000 = ( 1 + ... + 997 + 998 + 999 ) / 1000 = 499500/1000 = 4995/10

theo thứ tự 1,6/4=1 và 1/2,2,5/2 ,500

các bạn giải ra giup mik đi

a) \(A=1.2+2.3+3.4+...+999.1000\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+999.1000.\left(1001-998\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+999.1000.1001-998.999.1000\)

\(=999.1000.1001\)

\(A=\frac{999.1000.1001}{3}\)

b) \(B=1.3+3.5+5.7+...+999.1001\)

\(6B=1.3.6+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+999.1001.\left(1003-997\right)\)

\(=1.3.6+3.5.7-1.3.5+5.7.9-3.5.7+...+999.1001.1003-997.999.1003\)

\(=999.1001.1003+1.3\)

\(B=\frac{999.1001.1003+1.3}{6}\)

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\(\frac{2}{3}+\frac{1}{3}=1=\frac{2}{2}\)

\(\frac{3}{4}+\frac{2}{4}+\frac{1}{4}=\frac{6}{4}=\frac{3}{2}\);

\(\frac{4}{5}+\frac{3}{5}+\frac{2}{5}+\frac{1}{5}=2=\frac{4}{2}\)

;\(\frac{5}{6}+\frac{4}{6}+\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{15}{6}=\frac{5}{2}\)

Tổng quát:

\(\frac{n-1}{n}+\frac{n-2}{n}+...+\frac{2}{n}+\frac{1}{n}\)(\(n\in N\)) \(=\frac{n-1}{2}\)

Áp dụng:

\(\frac{999}{1000}+\frac{998}{1000}+\frac{997}{1000}+...+\frac{1}{1000}=\frac{999}{2}\).

Xem bài mình đúng không?

đây là bài tập cô tớ giao

\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+...+\frac{2016}{501}}{\frac{-1}{1.2}+\frac{-1}{3.4}+...+\frac{-1}{999.1000}}=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)

\(=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{500}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+....+\frac{1}{999}+\frac{1}{1000}\right)}=\frac{2016}{-1}=-2016\)

Vậy B = - 2016

Bạn Xyz cho mik hỏi ở phần mẫu số tại sao lại có -2*(1/2+1/4+...+1/1000) vậy? Nó ở đâu ra thế?