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=1/3-1/4+1/4-1/5+...+1/2005-1/2006
=1/3-1/2006
=2006/6018-3/6018
=2003/6018
Nếu đúng thì giữ lời hứa nhé
Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)
Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)
=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)
=> A < 1
Bài 1 :
36/1212 = 3/101
13/1313 = 1/101
3/101 + 1/101 = 4/101
Vậy 36/1212 + 13/1313 = 4/101.
Bài 2 :
A = 5/13 + 1/2 + -5/9 + -3/6 + 4/-9
A = 5/13 + 1/2 + -5/9 + -1/2 + -4/9
A = (1/2 + -1/2) + (-5/9 + -4/9) + 5/13
A = 0 + (-1) + 5/13
A = (-1) + 5/13 = -13/13 + 5/13 = 8/13.
Chúc bạn học giỏi nhé.
\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{2005\times2006}\) =\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2005}-\frac{1}{2006}\) =\(\frac{1}{3}-\frac{1}{2006}=\frac{2003}{6018}\)
a,\(\frac{5}{3}.\frac{3}{7}+\frac{5}{3}.\frac{5}{7}-\frac{5}{3}\)
=\(\frac{5}{3}.\left(\frac{3}{7}+\frac{5}{7}\right)-\frac{5}{3}\)
= \(\frac{5}{21}\)
B = \(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{53.56}\)
B = \(\frac{6-3}{3.6}+\frac{9-6}{6.9}+...+\frac{56-53}{53.56}\)
B = \(\frac{6}{3.6}-\frac{3}{3.6}+...+\frac{56}{53.56}-\frac{53}{53.56}\)
B = \(\frac{1}{3}-\frac{1}{6}+...+\frac{1}{53}-\frac{1}{56}\)
B = \(\frac{1}{3}-\frac{1}{56}\)
B = \(\frac{53}{168}\)
Ta có:
\(B=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.11}+...+\frac{3}{53.56}\)
\(=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{56}\)
\(=\frac{1}{3}-\frac{1}{56}=\frac{53}{168}\)
Vậy B=\(\frac{53}{168}\)