Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
x . 1/4 + x . 1/5 + x . 2 = 19,6
x . ( 1/4 + 1/5 + 2 ) = 19,6
x . 49/20 = 19,6
x = 19,6 : 49/20
x = 8
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2010}\right)\cdot x=2009\)
\(\frac{2009}{2010}\cdot x=2009\)
\(x=2009:\frac{2009}{2010}\)
\(x=2010\)
\(\Leftrightarrow10\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{x\times\left(x+1\right)}\right)=9\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=9\div10\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{9}{10}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{10}\)
\(\Rightarrow x+1=10\)
\(\Leftrightarrow x=9\)
Vậy x = 9
Áp dụng công thức \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\) ta đc:
\(\frac{2x+5}{x+5}=\frac{6}{4}\)( . là dấu nhân nha )
\(4\left(2x+5\right)=6\left(x+5\right)\)
\(8x+20=6x+30\)
\(2x=10\)
\(\Rightarrow x=5\)
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
\(\left(4.5-2\cdot x\right):\frac{3}{4}=1\frac{1}{3}\)
\(\left(4.5-2x\right):\frac{3}{4}=\frac{4}{3}\)
\(\left(4.5-2x\right)=\frac{4}{3}\cdot\frac{3}{4}\)
\(4.5-2x=1\)
\(2x=4.5-1\)
\(2x=3.5\)
\(x=3.5:2\)
\(x=1.75\)
(4,5 - 2 x X) : 3/4 = 1 1/3
(4,5 - 2 x X) = 1 1/3 x 3/4
(4,5 - 2 x X) = 1
2 x X = 4,5 - 1
2 x X = 3,5
X = 3,5 : 2
X = 1,75
chúc bạn học giỏi ^_^
tk mk nha !!!
=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 = \(\dfrac{502}{3}\)
\(x\times2+x\times\dfrac{1}{5}=1\dfrac{3}{5}\\ \Leftrightarrow x\times\left(2+\dfrac{1}{5}\right)=1\dfrac{3}{5}\\ \Leftrightarrow x\times\dfrac{12}{5}=\dfrac{8}{5}\\ \Rightarrow x=\dfrac{8}{5}:\dfrac{12}{5}=\dfrac{8}{12}=\dfrac{2}{3}\)
Đs...