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\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+....+1\right)=0\)
\(\Leftrightarrow x-2013=0\)(because 1/2012 +1/2011+...+1 luôn lớn hơn 0
\(\Leftrightarrow x=2013\)
Vậy ........
bạn bấm vào đúng 0 sẽ ra kết quả
mình làm bài này rồi
1.\(\frac{x+1}{2013}\)+\(\frac{x+2}{2012}\)=\(\frac{x+3}{2011}\)+\(\frac{x+4}{2010}\)
⇔\(\frac{x+1}{2013}\)+1+\(\frac{x+2}{2012}\)+1=\(\frac{x+3}{2011}\)+1+\(\frac{x+4}{2010}\)+1
⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)=\(\frac{x+2014}{2011}\)+\(\frac{x+2014}{2010}\)
⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)-\(\frac{x+2014}{2011}\)-\(\frac{x+2014}{2010}\)=0
⇔(x+2014)(\(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\))=0
Mà \(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\)≠0
⇔x+2014=0
⇔x=-2014
Vậy tập nghiệm của phương trình đã cho là:S={-2014}
2.\(\frac{3x+2}{4}\)+\(\frac{x+3}{2}\)=\(\frac{x-1}{3}\)-\(\frac{-x-1}{12}\)
⇔\(\frac{3\left(3x+2\right)}{12}\)+\(\frac{6\left(x+3\right)}{12}\)=\(\frac{4\left(x-1\right)}{12}\)+\(\frac{x+1}{12}\)
⇒9x+6+6x+18=4x-4+x+1
⇒15x+24=5x-3
⇒15x-5x=-3-24
⇒10x=-27
⇒ x=-\(\frac{27}{10}\)
Vậy tập nghiệm của phương trình đã cho là S={-\(\frac{27}{10}\)}
Ta có: Tử là:
B=\(\frac{1}{2013}+\frac{2}{2012}+...+\frac{2012}{2}+\left(1+1+...+1\right)\) (2013 số hạng 1)
=\(\left(\frac{1}{2013}+1\right)+\left(\frac{2}{2012}+1\right)+...+\left(\frac{2012}{2}+1\right)+\left(1\right)\)
=\(\frac{2014}{2013}+\frac{2014}{2012}+...+\frac{2014}{2}+\frac{2014}{2014}\)
=\(2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
=>A=\(\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2014
bấm vào đúng 0 sẽ ra kết quả, mình làm bài này rồi dễ lắm bạn ạ
\(\frac{x+1}{2012}+\frac{x+2}{2011}=\frac{x+3}{2010}+\frac{x+4}{2009}\)
\(\Leftrightarrow\frac{x+1}{2012}+1+\frac{x+2}{2011}+1=\frac{x+3}{2010}+1+\frac{x+4}{2009}+1\)
\(\Leftrightarrow\frac{x+2013}{2012}+\frac{x+2013}{2011}=\frac{x+2013}{2010}+\frac{x+2013}{2009}\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\Leftrightarrow x=-2013\)
\(\frac{x+1}{2012}+\frac{X+2}{2011}=\frac{X+3}{2010}+\frac{X+4}{2009}.\)
\(\Leftrightarrow\frac{X+1}{2012}+\frac{X+2}{2011}+2=\frac{X+3}{2010}+\frac{X+4}{2009}+2\)
\(\Leftrightarrow\frac{x+1}{2012}+1+\frac{x+2}{2011}+1=\frac{x+3}{2010}+1+\frac{x+4}{2009}+1\)
\(\Leftrightarrow\frac{x+2013}{2012}+\frac{x+2013}{2012}=\frac{x+2013}{2010}+\frac{x+2013}{2009}\)
\(\Leftrightarrow\left(x+2013\right).\left\{\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right\}=0\)
Mà \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}>0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
KL ; PT có Nghiệm \(S=\left\{-2013\right\}\)
\(a,⇔\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(⇔(x-23)(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27})=0\)
\(⇔x-23=0\) (vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\))
\(⇔x=23\)
\(b,⇔\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}+\frac{x+100}{95}=0\)
\(⇔(x+100)(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95})=0\)
\(⇔x+100=0\) (vì \(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}>0\))
\(⇔x=-100\)
\(c,⇔(\frac{x+1}{2012}+1)+(\frac{x+2}{2011}+1)=(\frac{x+3}{2010}+1)+(\frac{x+4}{2009}+1)\)
\(⇔\frac{x+2013}{2012}+\frac{x+2013}{2011}-\frac{x+2013}{2010}-\frac{x+2013}{2009}=0\)
\(⇔(x+2013)(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009})=0\)
\(⇔x+2013=0\) (vì \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}<0\))
\(⇔x=-2013\)
\(\frac{201-x}{99}+\frac{203}{97}=\frac{205}{95}+3\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)
Giúp mk với ạ
Ta có: \(\frac{x-1}{2012}+\frac{x-2}{2011}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+....+\frac{x-2012}{1}-1=2012-2012\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2011}+....+1\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=2013\)
Vậy x = 2013
PT đã cho tương đương với:
\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1+2010=2012\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
\(\Leftrightarrow x=2013\)