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a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
TL:
\(\frac{12}{100}\)= 0,12
\(\frac{5}{100}\)= 0,05
\(\frac{306}{1000}\)= 0,306
-HT-
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
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 = \(\dfrac{502}{3}\)
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
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
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
\(\dfrac{13+x}{20}\) = \(\dfrac{3}{4}\)
13 + \(x\) = 20 \(\times\) \(\dfrac{3}{4}\)
13 + \(x\) = 15
\(x\) = 15 - 13
\(x\) = 2
Cách khác :
\(\dfrac{13+x}{20}=\dfrac{3}{4}\)
\(\dfrac{13+x}{20}=\dfrac{15}{20}\)
\(13+x=15\)
\(x=15-13\)
\(x=2\)
bài 1:Thực hiện các phép tính sau
a. \(\dfrac{5}{3}\times\dfrac{2}{3}:\dfrac{1}{2}=\dfrac{5}{3}\times\dfrac{2}{3}\times\dfrac{2}{1}=\dfrac{5\times2\times2}{3\times3\times1}=\dfrac{20}{9}\)
b. \(\dfrac{2}{3}\times\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{2}{6}+\dfrac{1}{3}=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\)
c.\(\left(\dfrac{5}{7}-\dfrac{2}{5}\right):\dfrac{11}{7}=\left(\dfrac{5}{7}-\dfrac{2}{5}\right)\times\dfrac{7}{11}\)
\(=\left(\dfrac{5}{7}-\dfrac{2}{5}\right)\times\dfrac{7}{11}=\dfrac{5}{7}\times\dfrac{7}{11}-\dfrac{2}{5}\times\dfrac{7}{11}\)
\(=\dfrac{5}{11}-\dfrac{14}{55}=\dfrac{5\times5-14}{55}=\dfrac{11}{55}=\dfrac{1}{5}\)
bài 2:Viết mỗi phân số sau dưới dạng phân số thập phân:
\(\dfrac{5}{2}=\dfrac{25}{10};\dfrac{7}{4}=\dfrac{175}{100};\dfrac{31}{25}=\dfrac{124}{100};\dfrac{37}{50}=\dfrac{74}{100}\)
\(\dfrac{123}{125}=\dfrac{984}{1000};\dfrac{97}{20}=\dfrac{485}{100}\)
bài 3: Tìm xx
a. \(x\times\dfrac{1}{2}=\dfrac{5}{12}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{9}{12}=\dfrac{14}{12}=\dfrac{7}{6}\)
\(\Rightarrow x=\dfrac{7}{6}:\dfrac{1}{2}=\dfrac{7}{6}\times\dfrac{2}{1}=\dfrac{7}{3}\)
b.\(\dfrac{49}{8}:x=1+\dfrac{44}{5}=\dfrac{5}{5}+\dfrac{44}{5}=\dfrac{49}{5}\)
\(\Rightarrow x=\dfrac{49}{8}:\dfrac{49}{5}=\dfrac{49}{8}\times\dfrac{5}{49}=\dfrac{5}{8}\)
Đs...