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\(\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}+\dfrac{x+5}{2007}+\dfrac{x+2007}{5}=-5\)
Ta có:
\(\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}+1+\dfrac{x+5}{2007}+1+\dfrac{x+2007}{5}+1=0\)
\(=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}+\dfrac{x+2012}{2007}+\dfrac{x+2012}{5}=0\)
\(=\left(x+2012\right)\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{5}\right)=0\)
Mà \(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{5}\ne0\)
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
Vậy \(x=-2012\)
Chúc bạn học tốt!
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)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
Đặt D1 = \(\dfrac{2010}{1}\) + \(\dfrac{2009}{2}\) + \(\dfrac{2008}{3}\) + ... + \(\dfrac{1}{2010}\)
= 1 + ( 1+ \(\dfrac{2009}{2}\)) + ( 1+ \(\dfrac{2008}{3}\)) + ... + (1+\(\dfrac{1}{2010}\))
= \(\dfrac{2011}{2}\) + \(\dfrac{2011}{3}\)+ ... + \(\dfrac{2011}{2010}\) + \(\dfrac{2011}{2011}\)
= 2011. ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\))
Đặt D2 = \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\)
=> D = 2011
cho mk 1 tick nha
Ta có :
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì :
\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
Nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
Ta có: \(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì \(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)
hay A > B
Vậy A > B
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
\(B=\dfrac{2008+2009+2010}{2009+2010+2011}=\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)Ta có : \(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)\(=>\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}>\dfrac{2008+2009+2010}{2009+2010+2011}\)
Hay A > B
2, ta thấy:
\(\dfrac{2008}{2009}< \dfrac{2008}{2009+2010}\left(1\right)\)
\(\dfrac{2009}{2010}< \dfrac{2009}{2009+20010}\left(2\right)\)
từ (1) và (2) cộng vế với vế ta đc :\(\dfrac{2008}{2009}+\dfrac{2009}{20010}< \dfrac{2008}{2009+2010}+\dfrac{2009}{2009+2010}=\dfrac{2008+2009}{2009+2010}\)
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
\(\Rightarrow\left(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}\right)-\left(\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\right)=0\)
\(\Rightarrow\dfrac{x+3}{2007}-\dfrac{x+3}{2008}-\dfrac{x+3}{2010}+\dfrac{x+3}{2009}=0\)
\(\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\)
\(x+3=0\Rightarrow x=-3\)
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\\ \dfrac{x+3}{2007}-\dfrac{x+3}{2008}-\dfrac{x+3}{2010}+\dfrac{x+3}{2009}=0\\ \left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x+3=0\Rightarrow x=-3\)
\(\dfrac{x+1}{2011}+\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}=-4\)
\(\Rightarrow\dfrac{x+1}{2011}+1+\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}+1=0\)
\(\Rightarrow\dfrac{x+2012}{2011}+\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}=0\)
\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\ne0\)
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
Vậy x = -2012
Cảm ơn bạn nha