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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
30 tháng 1 2024
a; A = \(\dfrac{4026\times2014+4030}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2014+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2013\times2+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-4026+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2011\right)}{2013\times2016-2011}\)
A = 2
7 tháng 8 2017
\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
TP
7 tháng 8 2017
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)
13 tháng 5 2023
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 5 2023
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
3 tháng 9 2018
\(\frac{9}{16}:\frac{27}{8}=\frac{3^2:3^3}{2^4:2^3}=\frac{\frac{1}{3}}{2}=\frac{1}{6}\)
b tương tự nha
3 tháng 9 2018
\(\frac{9}{16}:\frac{27}{8}=\frac{9}{16}\cdot\frac{8}{21}=\frac{3\cdot3\cdot4\cdot2}{8\cdot2\cdot7\cdot3}=\frac{14}{3}\)
\(\frac{40}{7}:\frac{5}{14}=\frac{40}{7}\cdot\frac{14}{5}=\frac{4\cdot5\cdot2\cdot7\cdot2}{7\cdot5}=\frac{16}{1}=16\)
Chúc bạn học tốt
HQ
Hà Quang Minh
Giáo viên
28 tháng 1 2024
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
XO
4 tháng 6 2019
a) \(\frac{1}{2}\times x-3=6\)
=> \(\frac{1}{2}\times x=6+3\)
=> \(\frac{1}{2}\times x=9\)
=>\(x=9:\frac{1}{2}\)
=> \(x=18\)
b) \(2:x=\frac{2}{5}--\frac{1}{10}\)
=> \(2:x=\frac{2}{5}+\frac{1}{10}\)
=> \(2:x=\frac{1}{2}\)
=> \(x=2:\frac{1}{2}\)
=> \(x=4\)
c) \(25-\left(2\frac{1}{2}+x\right)=10\)
=> \(2\frac{1}{2}+x=25-10\)
=> \(\frac{5}{2}+x=15\)
=>\(x=15-\frac{5}{2}\)
=> \(x=\frac{25}{2}\)
d) \(\left(x-\frac{3}{4}\right)\times3-45:9=10\)
=> \(\left(x-\frac{3}{4}\right)\times3-5=10\)
=> \(\left(x-\frac{3}{4}\right)\times3=10+5\)
=> \(\left(x-\frac{3}{4}\right)\times3=15\)
=> \(\left(x-\frac{3}{4}\right)=15:3\)
=> \(\left(x-\frac{3}{4}\right)=5\)
=> \(x=5+\frac{3}{4}\)
=> \(x=\frac{23}{4}\)
4 tháng 6 2019
\(a,\frac{1}{2}.x-3=6\Rightarrow\frac{x}{2}=9\Rightarrow x=18\)
\(b,2:x=\frac{2}{5}-\frac{1}{10}\Rightarrow\frac{2}{x}=\frac{9}{10}\Rightarrow x=\frac{2.10}{9}=\frac{20}{9}\)
\(c,25-\left(2\frac{1}{2}+x\right)=10\Rightarrow25-\frac{5}{2}+x=10\Rightarrow x=10+\frac{5}{2}-25=-\frac{25}{2}\)
\(d,\left(x-\frac{3}{4}\right).3-45:9=10\Rightarrow\left(x-\frac{3}{4}\right).3-5=10\Rightarrow\left(x-\frac{3}{4}\right).3=15\Rightarrow x-\frac{3}{4}=5\Rightarrow x=\frac{23}{4}\)
\(\dfrac{15}{10}=\dfrac{3}{2}\)
\(\dfrac{9}{10}+\dfrac{6}{10}=\dfrac{9+6}{10}=\dfrac{15}{10}=\dfrac{3}{2}\)