Tính nhanh
2 phần 3.5 + 2 phần 5.7 + 2 phần 7.9 +...+ 2 phần 97.99
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A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
= \(1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
= \(2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\right)\)
= \(2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
= \(\frac{2}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)
= \(\frac{4}{13}\)
C = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)
= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\)
= \(\frac{9}{38}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
Ta có: \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2005.2006}\)
\(\Rightarrow N=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2005}-\frac{1}{2006}\)
\(=1-\frac{1}{2006}=\frac{2005}{2006}\)
\(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2015.2017}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
N = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/2005 - 1/2006
= 1/1 - 1/2006
= 2006/2006 - 1/2006
= 2005/2006
TÍNH NHANH
B=2 phần 1.3 + 2 phần 3.5 + 2 phần 5.7 +................+ 2 phần 99.101
(Giải thích rõ nha)
B=\(\frac{2}{1.3}+\frac{2}{3.5}+..........+\frac{2}{99.101}\)
B=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...........+\frac{1}{99}-\frac{1}{101}\)
B=\(1-\frac{1}{101}\)
B=\(\frac{100}{101}\)
A = 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2017. 2019
= ( 1 - 1/3 ) + ( 1/3 - 1/5 ) + ... + (1/2017 - 1/2019 )
= 1 - 1/2019
= 2018/2019
S = 1/31 + 1/32 +...+ 1/60
Ta có các phân số : 1/31, 1/32, ..., 1/59 đều lớn hơn 1/60
Nên S > 1/60 + 1/60 + 1/60 +...+ 1/60 ( có tất cả 30 phân số )
= 30/60 = 1/2
Vì 1/2 < 4/5 nên S <4/5
Vậy, chứng tỏ S < 4/5
Chúc bạn học tốt !
a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}=\frac{2017}{2018}\)
b, \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{2005}\right)\)
\(=\frac{1}{2}\cdot\frac{2004}{2005}=\frac{1002}{2005}\)
\(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\) Từ đó áp dụng tính câu a
\(\frac{2}{1.3}=\frac{1}{1}-\frac{1}{3}\) Áp dụng tính câu b
\(M=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(M=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(M=2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)
\(M=2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(M=\frac{1}{3}-\frac{1}{99}\)
\(M=\frac{32}{99}\)
M = 2 / 3.5 + 2 / 5.7 + 2 / 7.9 +...+2 / 97.99
M = 5 - 3 / 3 . 5 + 7 - 5 / 5 .7 + 9 - 7 / 7 . 9 +...+ 99 - 97 / 97 .99
M = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 +...+ 1/97 - 1/99
M = 1/3 - 1/99
M = 33 /99 - 1/99
M = 32/99
vậy M= 32/99
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{97.99}\)
\(=\frac{2}{2}.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{97.99}\right)\)
\(=1.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=1.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=1.\frac{33-1}{99}\)
\(=\frac{32}{99}\)
...................................TK CHO MK NHÉ.........................