NT
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Khách
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25 tháng 5 2018
a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{5}-\frac{1}{10}\)
\(=\frac{1}{10}\)
b) \(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{998.1000}\)
\(=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{998}-\frac{1}{1000}\)
\(=\frac{1}{10}-\frac{1}{1000}\)
\(=\frac{99}{1000}\)
c) \(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{69.90}\)
\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{89.90}\right)\)
\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{89}-\frac{1}{90}\right)\)
\(=4.\left(1-\frac{1}{90}\right)\)
\(=4.\frac{89}{90}\)
\(=\frac{178}{45}\)
_Chúc bạn học tốt_
TN
4 tháng 6 2019
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
chúc bn học tốt
TN
4 tháng 6 2019
\(1,27+2,77+4,27+5,77+...+31,27+32,47\)
\(=\left(1,27+32,77\right)+\left(2,77+31,27\right)+....+\left(16,27+17,77\right)\)
\(=34,04+34,04+....+34,04\)( 11 số hạng)
\(=34,04.11=374,44\)
chúc bn học tốt
PT
1
DD
Đoàn Đức Hà
Giáo viên
21 tháng 7 2021
\(A=\frac{1}{2\times3}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{98\times99}\)
\(=\frac{1}{6}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}+...+\frac{1}{98\times99}\)
\(=\frac{1}{6}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{99}\)
\(=\frac{1}{6}+\frac{1}{4}-\frac{1}{99}=\frac{161}{396}>\frac{160}{400}=\frac{2}{5}\)
19 tháng 9 2023
\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)+\left(\dfrac{5}{6}+\dfrac{19}{20}+...+\dfrac{2549}{2550}\right)\)
\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+..+\dfrac{1}{50\cdot51}\right)+\left(1-\dfrac{1}{2\cdot3}\right)+\left(1-\dfrac{1}{3\cdot4}\right)+...+\left(1-\dfrac{1}{50\cdot51}\right)\)
\(B=\left(1+1+...+1\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\)
\(B=1\cdot49=49\) (vì có (50 - 2) : 1 + 1 = 49 số hạng 1)
24 tháng 4 2021
Ta có: \(C=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}\)
\(\Leftrightarrow C=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\right)\)
\(\Leftrightarrow C=2\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(\Leftrightarrow C=2\left(1-\dfrac{1}{7}\right)=\dfrac{2.6}{7}=\dfrac{12}{7}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
3 tháng 7 2015
Bài 1:
Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)
\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)
\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)
\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
\(A=\frac{2x19}{20}=\frac{19}{10}\)
Bài 2:
Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)
Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)
\(Bx100=\frac{9}{10}x100=90\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)
=> \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)
Bài làm:
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{9.10}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{10}\right)=2.\frac{4}{10}=\frac{4}{5}\)