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\(x.\frac{1}{2}+\frac{1}{4}x=\frac{4}{5}.3\)
\(\frac{3}{4}x=\frac{12}{5}\)
\(x=\frac{16}{5}\)
Vậy \(x=\frac{16}{5}\)
Suy ra X x 1/2 + 1/4X = 12/5
Suy ra X x ( 2/4 + 1/4)=12/5
Suy ra X x 3/4 = 12/5
Suy ra X = 12/5 : 3/4
Suy ra X = 48/15
1/2 x y + 1/3 x y + 1/4 x y = 11/4
y x ( 1/2 + 1/3 + 1/4 ) = 11/4
y x 13/12 = 11/4
y = 11/4 : 13/12
y = 11/4 x 12/13
y = 33/13
^^ Học tốt!
\(\frac{1}{2}\times Y+\frac{1}{3}\times Y+\frac{1}{4}\times Y=\frac{11}{4}\)
\(3Y=\frac{11}{4}\div\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
\(3Y=\frac{33}{13}\)
\(Y=\frac{33}{13}\div3\)
\(Y=\frac{11}{13}\)
~ Chúc bạn học tốt
=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 = \(\dfrac{502}{3}\)
(3/10+4/5x1/2):1/8/9-1/1/3
=(3/10+4/10):17/9-4/3
=7/10:17/9-4/3
=63/170-4/3
=-491/510
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{1}-\frac{1}{8}\)
\(=\frac{7}{8}\)'
\(A=1\dfrac{1}{2}\times1\dfrac{1}{3}\times1\dfrac{1}{4}\times....\times1\dfrac{1}{2020}\times1\dfrac{1}{2021}\\ =\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times....\times\dfrac{2022}{2021}\\ =\dfrac{3\times4\times5\times6\times.....\times2022}{2\times3\times4\times5\times....\times2021}\\ =\dfrac{2022}{2}=1011\)