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`(x-1)/2013+(x-2)/2012+(x-3)/2011=(x-4)/2010+(x-5)/2009 +(x-6)/2008`
`<=> ((x-1)/2013-1)+((x-2)/2012-1)+((x-3)/2011-1)=( (x-4)/2010-1)+((x-5)/2009-1)+((x-6)/2008-1)`
`<=> (x-2014)/2013 +(x-2014)/2012+(x-2014)/2011=(x-2014)/2010+(x-2014)/2009+(x-2014)/2008`
`<=> x-2014=0` (Vì `1/2013+1/2012+1/2011-1/2010-1/2009-1/2008 \ne 0`)
`<=>x=2014`
Vậy `S={2014}`.
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2013}-1\right)+\left(\dfrac{x-2}{2012}-1\right)+\left(\dfrac{x-3}{2011}-1\right)=\left(\dfrac{x-4}{2010}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-6}{2008}-1\right)\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
\(\Leftrightarrow\left(x-2014\right).A=0\)
\(\text{Vì A }\ne0\)
\(\Rightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2014\right\}\)
Bài của bạn nè bạn gái!
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{1012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
mà \(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{10}{2008}\ne0\)
\(\Rightarrow x-2014=0\Rightarrow x=2014\)
vậy x=2014
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}+1+\dfrac{x-2}{2012}+1+\dfrac{x-3}{2011}+1-\dfrac{x-4}{2010}+1-\dfrac{x-5}{2009}+1-\dfrac{x-6}{2008}+1=0\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
Vậy PT có nghiệm là \(x=2014\)
a) \(\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\)
\(\Rightarrow\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}+4=0\)
\(\Rightarrow\left(\dfrac{x+5}{2010}+1\right)+\left(\dfrac{x+4}{2011}+1\right)+\left(\dfrac{x+3}{2012}+1\right)+\left(\dfrac{x+2}{2013}+1\right)=0\)
\(\Rightarrow\left(\dfrac{x+5}{2010}+\dfrac{2010}{2010}\right)+\left(\dfrac{x+4}{2011}+\dfrac{2011}{2011}\right)+\left(\dfrac{x+3}{2012}+\dfrac{2012}{2012}\right)+\left(\dfrac{x+2}{2013}+\dfrac{2013}{2013}\right)=0\)
\(\Rightarrow\dfrac{x+5+2010}{2010}+\dfrac{x+4+2011}{2011}+\dfrac{x+3+2012}{2012}+\dfrac{x+2+2013}{2013}=0\)
\(\Rightarrow\dfrac{x+2015}{2010}+\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}+\dfrac{x+2015}{2013}=0\)
\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Mà \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\ne0\)
\(\Rightarrow x+2015=0\)
\(\Rightarrow x=-2015\)
b) \(\dfrac{x-22}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Rightarrow\dfrac{x-22}{77}+\dfrac{x-11}{78}-2=\dfrac{x-74}{15}+\dfrac{x-73}{16}-2\)
\(\Rightarrow\left(\dfrac{x-22}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)=\left(\dfrac{x-74}{15}-1\right)+\left(\dfrac{x-73}{16}-1\right)\)
\(\Rightarrow\left(\dfrac{x-22}{77}-\dfrac{77}{77}\right)+\left(\dfrac{x-11}{78}-\dfrac{78}{78}\right)=\left(\dfrac{x-74}{15}-\dfrac{15}{15}\right)+\left(\dfrac{x-73}{16}-\dfrac{16}{16}\right)\)
\(\Rightarrow\dfrac{x-22-77}{77}+\dfrac{x-11-78}{78}=\dfrac{x-74-15}{15}+\dfrac{x-73-16}{16}\)
\(\Rightarrow\dfrac{x-99}{77}+\dfrac{x-99}{78}=\dfrac{x-99}{15}+\dfrac{x-99}{16}\)
\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{77}+\dfrac{1}{78}\right)=\left(x-99\right)\left(\dfrac{1}{15}+\dfrac{1}{16}\right)\)
\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{77}+\dfrac{1}{78}\right)-\left(x-99\right)\left(\dfrac{1}{15}+\dfrac{1}{16}\right)=0\)
\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
Mà \(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\ne0\)
\(\Rightarrow x-99=0\)
\(\Rightarrow x=99\)
\(\Rightarrow\frac{x}{2010}+\frac{x+1}{2011}+\frac{x+2}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}-5=0\)
\(\left(\frac{x}{2010}-1\right)+\left(\frac{x+1}{2011}-1\right)+\left(\frac{x+2}{2012}-1\right)\)\(+\left(\frac{x+3}{2013}-1\right)+\left(\frac{x+4}{2014}-1\right)=0\)
\(\frac{x-2010}{2010}+\frac{x-2010}{2011}+\frac{x-2010}{2012}+\frac{x-2010}{2013}+\frac{x-2010}{2014}=0\)
\(\left(x-2010\right).\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)=0\)
mà \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)
Vậy x=-2010
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
<=>\(\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+\dfrac{x-3}{2010}-1+...+\dfrac{x-2012}{1}-1=0\)
<=>\(\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)
<=>\(\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+1\right)=0\)
do 1/2012+1/2011....+1 khác 0 =>x-2013=0<=>x=2013
vậy..........................
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
\(\left(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}\right)-2012=0\)
\(\Rightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)
\(\Rightarrow x-2013\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\right)=0\)
Vì \(x-2013\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\right)=0\)nên x - 2013 hoặc \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\) = 0. Nhưng \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\ne0\) nên x - 2013 = 0. Vì vậy x = 2013.
Vậy...
Tìm x:
\(\dfrac{x}{2010}+\dfrac{x+1}{2011}+\dfrac{x+2}{2012}+\dfrac{x+3}{2013}+\dfrac{x+4}{2014}=5\)
\(\dfrac{x}{2010}+\dfrac{x+1}{2011}+\dfrac{x+2}{2012}+\dfrac{x+3}{2013}+\dfrac{x+4}{2014}=5\)
\(\Leftrightarrow\left(\dfrac{x}{2010}-1\right)+\left(\dfrac{x+1}{2011}-1\right)+\left(\dfrac{x+2}{2012}-1\right)+\left(\dfrac{x+3}{2013}-1\right)+\left(\dfrac{x+4}{2014}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2010}{2010}+\dfrac{x-2010}{2011}+\dfrac{x-2010}{2012}+\dfrac{x-2010}{2013}+\dfrac{x-2010}{2014}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)=0\)
\(\Leftrightarrow x=2010\)
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
\(\Leftrightarrow\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+...+\dfrac{x-2012}{1}-1=0\)
\(\Leftrightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+...+\dfrac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}\right)=0\)
Dễ thấy: \(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}>0\)
\(\Rightarrow x-2013=0\Rightarrow x=2013\)
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