tim x,biet |x+\(\frac{3}{4}\)|=x
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\(\frac{x-3}{2013}+\frac{x-4}{2012}=\frac{x-5}{2011}+\frac{x-6}{2010}\)
\(\Leftrightarrow\frac{x-3-2013}{2013}+\frac{x-2-2012}{2012}=\frac{x-5-2011}{2011}+\frac{x-6-2010}{2010}\)(mỗi vế trừ đi 2)
\(\Leftrightarrow\frac{x-2016}{2013}+\frac{x-2016}{2012}-\frac{x-2016}{2011}-\frac{x-2016}{2010}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Mà \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)
\(\Rightarrow x-2016=0\Leftrightarrow x=2016\)
Cộng mỗi vế cho 1
Ta có: \(\frac{x-3-2013}{2013}+\frac{x-4-2012}{2012}=\frac{x-5-2011}{2011}+\frac{x-6-2010}{2010}\)
\(=>\left(\frac{x-2016}{2013}+\frac{x-2016}{2012}\right)-\left(\frac{x-2016}{2011}+\frac{x-2016}{2010}\right)=0\)
\(=>\left(x-2016\right).\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)\)
Mà \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\ne0\)
\(=>x-2016=0\\ =>x=2016\)
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Ta có :
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)
\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)
\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)
Nên \(x+50=0\)
\(\Rightarrow\)\(x=-50\)
Vậy \(x=-50\)
Chúc bạn học tốt ~
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\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\)
\(x=0-2004\)
\(x=-2004\)
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Ta có : \(\frac{x+2}{198}+\frac{x+3}{197}=\frac{x+4}{196}+\frac{x+5}{195}\)
=> \(\left(\frac{x+2}{198}+1\right)+\left(\frac{x+3}{197}+1\right)=\left(\frac{x+4}{196}+1\right)+\left(\frac{x+5}{195}+1\right)\)
=> \(\frac{x+2+198}{198}+\frac{x+3+197}{197}=\frac{x+4+196}{196}+\frac{x+5+195}{195}\)
=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)
=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)
=> \(\left(x+200\right)\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)=0\)
Ta có : \(\frac{1}{198}+\frac{1}{197}\ne\frac{1}{196}+\frac{1}{195}\) => \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\ne0\)
=> x + 200 = 0
=> x = -200
<=> (\(\frac{x+2}{198}\)+1) +(\(\frac{x+3}{197}\)+1) =(\(\frac{x+4}{196}\)+1) +(\(\frac{x+5}{195}\)+1)
<=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)
<=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)
<=> \(\left(x+200\right)\cdot\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)\)=0
Vì \(\frac{1}{195}>\frac{1}{196}>\frac{1}{197}>\frac{1}{198}\)
<=> \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\) khác 0
<=> \(x+200=0\)
<=> x =
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=> x+3/4 =x hoặc x+3/4 = -x
=> 0 = 3/4 ( loại ) hoặc 2x = 3/4
=> x = 3/4 :2 = 3/8