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\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}=\frac{x-4}{2011}\)
\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}-\frac{x-4}{2011}=0\)
\(\left(\frac{x-1}{2014}-1\right)+\left(\frac{x-2}{2013}-1\right)-\left(\frac{x-3}{2012}-1\right)-\left(\frac{x-4}{2011}-1\right)=0\)
\(\frac{x-2015}{2014}+\frac{x-2015}{2013}-\frac{x-2015}{2012}-\frac{x-2015}{2011}=0\)
\(\left(x-2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\)
\(\Rightarrow x-2015=0\)
\(x=0+2015\)
\(x=2015\)
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)
\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)
Vì \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
=> \(x+2014=0\)
\(x=0-2014\)
\(x=-2014\)
a)
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\\ \Leftrightarrow2^x.1+2^x.2+2^x.2^2+2^x.2^3=120\\ \Leftrightarrow2^x\left(1+2+2^2+2^3\right)=120\\ \Leftrightarrow2^x=8=2^3\\ \Rightarrow x=3\)
b)
\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}=\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\\ \Leftrightarrow\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1=\dfrac{x+2}{2013}+1+\dfrac{x+1}{2014}+1\\ \Leftrightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}=\dfrac{x+2015}{2013}+\dfrac{x+2015}{2014}\\ \Leftrightarrow\left(x+2015\right).\dfrac{1}{2011}+\left(x+2015\right).\dfrac{1}{2012}-\left(x+2015\right).\dfrac{1}{2013}-\left(x+2015\right).\dfrac{1}{2014}=0\\ \Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\\ \Rightarrow x+2015=0\Leftrightarrow x=-2015\)
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
`=> (x+2014) (1/2010 + 1/2011-1/2012-1/2013)=0`
`=> x+2014=0` ( vì `1/2010 + 1/2011-1/2012-1/2013≠0 )`
`=>x=-2014`
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{2}{2012}\right)+\left(1+\dfrac{1}{2013}\right)+1\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}+\dfrac{2014}{2014}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)
\(\Leftrightarrow x=\dfrac{2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}\)
\(\Leftrightarrow x=2014\)
Vậy \(x=2014\)
\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{1}{2013}\\ =\dfrac{2012}{2}+1+\dfrac{2011}{3}+1+...+\dfrac{1}{2013}+1+1\\ =\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}+\dfrac{2014}{2014}\\ =2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\)
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\\ x=2014\)
Vậy x = 2014
\(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)
\(\Leftrightarrow\dfrac{x+4}{2012}+1+\dfrac{x+3}{2013}+1=\dfrac{x+2}{2014}+1+\dfrac{x+1}{2015}\)
\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}=\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\)
\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}-\left(\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\right)=0\)
\(\Leftrightarrow x+2016.\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\right)\)
Vì \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
Vậy \(x=-2016\) tại biểu thức \(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)
Theo đề ta có: x+4/2012+x+3/2013=x+2/2014+x+1/2015
=>x+4/2012+x+3/2013-x+2/2014+x+1/2015=0
=>( x+4/2012+1)+(x+3/2013+1)-(x+2/2014+1)+(x+1/2015+1)
=>x+2016/2012+x+2016/2013-x+2016/2014-x+2016/2015=0
=>x+2016.(1/2012+1/2013-1/2014-1/2015)=0
Do 1/2012+1/2013-1/2014-1/2015>0
nên x+2016=0
=>x=-2016
Vậy x=-2016
Ta có : \(\dfrac{x-3}{2015}+\dfrac{x-4}{2014}+\dfrac{x-5}{2013}+\dfrac{x-6}{2012}=4\)
\(\dfrac{x-3}{2015}+\dfrac{x-4}{2014}+\dfrac{x-5}{2013}+\dfrac{x-6}{2012}-4=0\)
\(\dfrac{x-3}{2015}-1+\dfrac{x-4}{2014}-1+\dfrac{x-5}{2013}-1+\dfrac{x-6}{2012}-1=0\)
\(\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}+\dfrac{x-2018}{2013}+\dfrac{x-2018}{2012}=0\)
\(\left(x-2018\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}\right)=0\)
Vì \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}>0\)
=> x - 2018 = 0
x = 0 + 2018
x = 2018
Vậy x = 2018
a) \(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\) \(\Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=-2\\\dfrac{x}{2}-\dfrac{1}{3}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}=\dfrac{-5}{3}\\\dfrac{x}{2}=\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{14}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{-10}{3}\) hoặc \(x=\dfrac{14}{3}\) thì thỏa mãn đề bài.
b) \(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\) \(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\) \(\Rightarrow\dfrac{x+4+2010}{2010}+\dfrac{x+3+2011}{2011}=\dfrac{x+2+2012}{2012}+\dfrac{x+1+2013}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\) \(\Rightarrow\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\) \(\Rightarrow x+2014=0\) \(\Rightarrow x=-2014\)
Vậy \(x=-2014\) thì thỏa mãn đề bài.
c) \(3^{x+2}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1+1}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}\times3+4\times3^{x+1}=7\times3^6\) \(\Rightarrow\left(3+4\right)\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}=3^6\) \(\Rightarrow x+1=6\) \(\Rightarrow x=5\)
Vậy \(x=5\) thì thỏa mãn đề bài.
a)
\(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\\ \Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=2\\\dfrac{x}{2}-\dfrac{1}{3}=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{1}{3}+2\\\dfrac{x}{2}=\dfrac{1}{3}-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{7}{3}\\\dfrac{x}{2}=\dfrac{-5}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{3}.2\\x=\dfrac{-5}{3}.2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\Rightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)
mà \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
=> x + 2014 = 0
=> x = -2014
vậy x = -2014
c)\(3^{x+2}+4.3^{x+1}=7.3^6\)
\(\Rightarrow3^{x+1}.3+4.3^{x+1}=7.3^6\\ \Rightarrow3^{x+1}\left(3+4\right)=7.3^6\\ \Rightarrow3^{x+1}.7=7.3^6\\ \Rightarrow3^{x+1}=3^6\\ \Rightarrow x+1=6\\ x=6-1\\ x=5\)
vậy x = 5
\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}=\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\)
\(\Rightarrow\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1=\dfrac{x+2}{2013}+1+\dfrac{x+1}{2014}+1\)
\(\Rightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}-\dfrac{x+2015}{2013}-\dfrac{x+2015}{2014}=0\)
\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\ne0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
Vậy x = -2015
\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\)
\(\Rightarrow\dfrac{x+4}{2011}+\dfrac{x+3}{2012}-\dfrac{x+2}{2013}-\dfrac{x+1}{2014}=0\)
\(\Rightarrow\)\(\left(\dfrac{x+4}{2011}+1\right)+\left(\dfrac{x+3}{2012}+1\right)-\left(\dfrac{x+2}{2013}+1\right)-\left(\dfrac{x+1}{2014}+1\right)=0\)\(\Rightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}-\dfrac{x+2015}{2013}-\dfrac{x+2015}{2014}=0\)
\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)