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a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)
b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)
Sửa lại đề:
\(M=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{49.50}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-..........-\dfrac{1}{49}-\dfrac{1}{50}\)
\(M=1-\dfrac{1}{50}\)
\(M=\dfrac{50}{50}-\dfrac{1}{50}\)
\(M=\dfrac{49}{50}\)
Đề bài là thu gọn / tính giá trị biểu thức nhé chứ không phải là So sánh , thiếu dữ kiện kìa
A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)
B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)
Bài 2:
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)
Vậy A < 2
Bài 3:
D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)
\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)
Bài 4:
A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)
\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)
\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
Ta có: \(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}\)
\(=\dfrac{a+1-a}{a\left(a+1\right)}\)
\(=\dfrac{1}{a\left(a+1\right)}\) (đpcm)
Ta được:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+...-\dfrac{1}{100}\) \(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
a, A = 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/2017 - 1/2018
A = 1 - 1/2018 = 2017/2018
b, B = 5/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2016 -1/2018)
B= 5/2 . ( 1/2 - 1/ 2018 )
B = 504/1009
c, C = 1/3.6 + 1/ 6.9 + 1/ 9.12 + ... + 1/ 30.33
C= 1/3 - 1/6 + 1/6 - 1/ 9 + 1/9 - 1/12 + ... + 1/30 - 1/33
C = 1/3 - 1/33
C= 10/33
phan B mk quên nhân với 5/2
lấy 5/2 . 504/1009 = 1260/1009
Ta có: \(M=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(\Rightarrow M\) \(\)\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow M=1-\dfrac{1}{50}< 1\)
Vậy M < 1.
M=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=1-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}.\)
Vậy M=\(\dfrac{49}{50}\)
*Trước dấu = là 1 chữ M
\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)
= \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{8}\) MSC: 8
= \(\dfrac{4}{8}\) + \(\dfrac{1}{8}\)
= \(\dfrac{5}{8}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
= \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
= \(\dfrac{1}{2}-\dfrac{1}{8}\)
=\(\dfrac{4}{8}-\dfrac{1}{8}\)
=\(\dfrac{3}{8}\)
a: \(A=\left(\dfrac{-3}{4}+\dfrac{-2}{9}-\dfrac{1}{36}\right)+\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{3}{5}\right)+\dfrac{1}{57}\)
\(=\dfrac{-27-8-1}{36}+\dfrac{5+1+9}{15}+\dfrac{1}{57}\)
=1/57
b: \(B=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}-\dfrac{5}{7}-\dfrac{3}{35}\right)+\dfrac{1}{41}\)
\(=\dfrac{3+1+2}{6}+\dfrac{-7-25-3}{35}+\dfrac{1}{41}\)
=1/41
c: \(C=\left(\dfrac{-1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{107}\)
=1-1+1/107
=1/107
Giải: Ta có: \(\dfrac{1}{2.3}\)=\(\dfrac{1}{2}-\dfrac{1}{3}\) ; \(\dfrac{1}{2}-\dfrac{1}{3}\)(giữ nguyên)
Vì \(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{1}{3}\) nên \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
Vậy \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)