\(\frac{1}{1000.1998}+\frac{1}{1001.1997}+...+\frac{1}{1998.1000}\)
K
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Những câu hỏi liên quan
17 tháng 6 2019
A = 1/(1.2) + 1/(3.4) + 1/(5.6) +....+ 1/(1997.1998) =
(1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 1997 - 1 / 1998) =
(1 + 1 / 2 + 1 / 3 + ... + 1998) - 2(1 / 2 + 1 / 4 + ... + 1 / 1998) =
(1 + 1 / 2 + 1 / 3 + ... + 1998) - (1 + 1 / 2 + ... + 1 / 999) =
1 / 1000 + 1 / 1001 + ... + 1 / 1998
2A = (1 / 1000 + 1 / 1001 + ... + 1 / 1998) + (1 / 1998 + 1 / 1997 + ... + 1 / 1000) =
(1 / 1000 + 1 / 1998) + (1 / 1001 + 1 / 1997) + ... + (1 / 1998 + 1 / 1000) =
2998*[1 / (1000*1998) + 1 / (1001*1997) + ... + 1 / (1998*1000)] = 2998B
=> A / B = 1499 nguyên
17 tháng 6 2019
A = (1/1.2) + (1/3.4) + (1/5.6) +....+ ( 1/1997.1998)
ta có
1/1*2 = 1 - 1/2
1/3*4 = 1/3 - 1/4
...
1/1997*1998 = 1/1007 - 1/1998
bạn gộp lại tự giải tiếp nha
NH
14 tháng 3 2018
Giup tui voi !!!!!!!!!!!!!!!!!!!!!!!!!!! Mai phai nop roi !!!!!!!!!!!!!!!!!!!
NT
27 tháng 2 2018
\(A=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}+\frac{1}{8}.\frac{1}{9}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(A=\frac{1}{2}-\frac{1}{9}\)
\(A=\frac{7}{18}\)
Vậy \(A=\frac{7}{18}\)
27 tháng 2 2018
A = 1/2.3 + 1/3.4 + ..... +1/8.9
= 1/2 - 1/3 + 1/3 - 1/4 + ........ + 1/8 - 1/9
= 1/2 - 1/9
= 7/18
Tk mk nha
\(\frac{1}{1000.1998}+\frac{1}{1001.1997}+...+\frac{1}{1998+1000}\)
\(S=\frac{1}{1000.1998}+\frac{1}{1001.1997}+...+\frac{1}{1998.1000}\)
\(=\frac{1}{2998}\left(\frac{1000+1998}{1000.1998}+\frac{1001+1997}{1001.1997}+...+\frac{1998+1000}{1998.1000}\right)\)
\(=\frac{1}{2998}\left(\frac{1}{1000}+\frac{1}{1998}+\frac{1}{1001}+\frac{1}{1997}+...+\frac{1}{1998}+\frac{1}{1000}\right)\)
\(=\frac{2}{2998}\left(\frac{1}{1000}+\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{1998}\right)\)
\(=\frac{1}{1499}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1998}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}\right)\right]\)
\(=\frac{1}{1499}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1998}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1998}\right)\right]\)
\(=\frac{1}{1499}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1997}-\frac{1}{1998}\right)\)
\(=\frac{1}{1499}\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{1997.1998}\right)\)
$\frac{1}{1000.1998}$ + $\frac{1}{1001.1997}$ +...+ $\frac{1}{1998.1000}$ câu hỏi 198346 - hoidap247.com