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H
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 4 2023
\(\dfrac{13+x}{20}\) = \(\dfrac{3}{4}\)
13 + \(x\) = 20 \(\times\) \(\dfrac{3}{4}\)
13 + \(x\) = 15
\(x\) = 15 - 13
\(x\) = 2
29 tháng 4 2023
Cách khác :
\(\dfrac{13+x}{20}=\dfrac{3}{4}\)
\(\dfrac{13+x}{20}=\dfrac{15}{20}\)
\(13+x=15\)
\(x=15-13\)
\(x=2\)
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
21 tháng 4 2021
a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
21 tháng 4 2021
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 1 2024
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 1 2024
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
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 = \(\dfrac{502}{3}\)
25 tháng 9 2020
\(\frac{8}{9}+\frac{1}{3}=\frac{8}{9}+\frac{3}{9}=\frac{11}{9}\)
25 tháng 9 2020
\(\frac{8}{9}+\frac{1}{3}=\frac{8}{9}+\frac{3}{9}=\frac{11}{9}\)
\(\frac{24}{15}-\frac{20}{25}=\frac{24}{15}-\frac{4}{5}=\frac{24}{15}-\frac{12}{15}=\frac{12}{15}\)
\(3\frac{1}{6}\times2\frac{3}{5}=\frac{19}{6}\times\frac{13}{5}=\frac{247}{30}\)
\(2\frac{1}{10}\div2\frac{2}{5}=\frac{21}{10}\div\frac{12}{5}=\frac{21}{10}\times\frac{5}{12}=\frac{7}{8}\)
MA
14 tháng 10 2016
a ) \(1\frac{1}{2}+2\frac{1}{3}+3\frac{1}{6}-5\)
\(=\frac{3}{2}+\frac{7}{3}+\frac{19}{6}-\frac{5}{1}\)
\(=\frac{9}{6}+\frac{14}{6}+\frac{19}{6}-\frac{30}{6}\)
\(=\frac{23}{6}+\frac{19}{6}-\frac{30}{6}\)
\(=\frac{42}{6}-\frac{30}{6}\)
\(=\frac{12}{6}=2\)
b ) \(2\frac{2}{3}\times3\frac{3}{4}\div4\frac{4}{5}\)
\(=\frac{8}{3}\times\frac{15}{4}\div\frac{24}{5}\)
\(=\frac{120}{12}\div\frac{24}{5}\)
\(=\frac{120}{12}\times\frac{5}{24}\)
\(=\frac{600}{288}=\frac{25}{12}\)
c ) \(4\frac{1}{5}+5\frac{1}{3}-2\frac{2}{3}\times3\frac{1}{5}+\frac{9}{25}\div\frac{9}{20}\)
\(=\frac{21}{5}+\frac{16}{3}-\frac{8}{3}\times\frac{16}{5}+\frac{9}{25}\div\frac{9}{20}\)
\(=\frac{63}{15}+\frac{80}{15}-\frac{128}{15}+\frac{9}{25}\times\frac{20}{9}\)
\(=\frac{143}{15}-\frac{128}{15}+\frac{180}{225}\)
\(=\frac{15}{15}+\frac{12}{15}\)
\(=\frac{27}{15}=\frac{9}{5}\)
\(20\%+\dfrac{2}{3}+1,8+2\dfrac{1}{3}\\ =\dfrac{1}{5}+\dfrac{2}{3}+\dfrac{9}{5}+\dfrac{7}{3}\\ =\left(\dfrac{1}{5}+\dfrac{9}{5}\right)+\left(\dfrac{2}{3}+\dfrac{7}{3}\right)\\ =\dfrac{10}{5}+\dfrac{9}{3}\\ =2+3\\ =5\)