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\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2011}{2013}\)
\(\Rightarrow2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2011}{2013}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2011}{4026}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2011}{4026}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
Chúc hok dốt!
a/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^9}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{10}}\)
b/ \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+.......+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Leftrightarrow3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+......+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{101}{1540}.3\)
\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+......+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+.....+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Leftrightarrow x+3=308\)
\(\Leftrightarrow x=305\)
Vậy ..
c/ \(1+\dfrac{1}{3}+\dfrac{1}{6}+........+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2007}{2009}\)
\(\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+.......+\dfrac{1}{x\left(x+1\right):2}\right)=\dfrac{4016}{2009}.\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)
\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2009}\)
\(\Leftrightarrow x+1=2009\)
\(\Leftrightarrow x=2008\)
Vậy ..
bài 1:
A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
ta thấy 2A=\(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^9}\)
=>2A-A=\(1-\dfrac{1}{2^{10}}=\dfrac{1023}{1024}\)
Lời giải:
Ta có:
\(\frac{1}{x(x+1):2}=\frac{2}{x(x+1)}=2.\frac{(x+1)-x}{x(x+1)}=2\left(\frac{1}{x}-\frac{1}{x+1}\right)\)
Do đó:
\(\frac{1}{3}=\frac{1}{2.3:2}=2\left(\frac{1}{2}-\frac{1}{3}\right)\)
\(\frac{1}{6}=\frac{1}{3.4:2}=2\left(\frac{1}{3}-\frac{1}{4}\right)\)
\(\frac{1}{10}=\frac{1}{4.5:2}=2\left(\frac{1}{4}-\frac{1}{5}\right)\)
.......
\(\frac{1}{x(x+1):2}=2\left(\frac{1}{x}-\frac{1}{x+1}\right)\)
Cộng theo vế:
\(\text{VT}=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{x}-\frac{1}{x+1}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)\) \(=1-\frac{2}{x+1}\)
Mà \(\text{VT}=\frac{2009}{2011}\Rightarrow 1-\frac{2}{x+1}=\frac{2009}{2011}\Rightarrow x=2010\)
Mình có cách giải khác:
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{2009}{2011}\)
=\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2009}{4022}\)
=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2009}{4022}\)
=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2009}{4022}_{ }\)
=\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2009}{4022}\)
➩ \(\dfrac{1}{x+1}=\dfrac{1}{2011}\)
➩ \(x=2011-1=2010\)
=> \(\dfrac{2}{6}\)+\(\dfrac{2}{12}\)+\(\dfrac{2}{20}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\)=\(\dfrac{2011}{2013}\)
=> 2.(\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{2011}{2013}\)
=> 2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{2011}{2013}\)
=> 2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\)) =\(\dfrac{2011}{2013}\)
=>\(\dfrac{x+1-2}{2.\left(x+1\right)}\)=\(\dfrac{2011}{2013}\)
=> \(\dfrac{x-1}{x+1}\)=\(\dfrac{2011}{2013}\)
=> 2013.(x-1) = 2011.(x+1)
=> 2013x-2013= 2011x+2011
=> 2013x -2011x= 2013+2011
=> 2x= 4024
=> x= 2012
Chúc bạn học tốt!Tick cho mk nhé
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
a, (x + 1) + (x + 4) + ... + (x + 28) = 155
x + 1 + x + 4 + ... + x + 28 = 155
(x + x + x + ... + x) + (1 + 4 + ... + 28) = 155
x . 10 + 145 = 155
x . 10 = 155 - 145
x . 10 = 10
x = 10 : 10
x = 1
\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+2\right)}=1\dfrac{2009}{2011}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{4020}{4022}\)
\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{4020}{4022}\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4020}{4022}\)
\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{4020}{4022}\)
\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{2011}\)
\(\Rightarrow x+2=2011\Rightarrow x=2009\)
Vậy x = 2009