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Lời giải:
Ta có:
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow \left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow \frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow (x-2013)\left(\frac{1}{2012}+\frac{1}{2011}+...+1\right)=0\)
Dễ thấy \(\frac{1}{2012}+\frac{1}{2011}+...+1\neq 0\Rightarrow x-2013=0\)
\(\Leftrightarrow x=2013\)
Vậy PT có nghiệm \(x=2013\)
\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
\(\Leftrightarrow\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-2014}{3}\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
Vậy x = 2014.
1, \(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\\ \\ < =>\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\\ \\ < =>\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\\ \\ < =>\left(x-2014\right).\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\\ \\ < =>x-2014=0< =>x=2014\)
2, \(x^2+1=x\\ \\ < =>x^2-x+1=0\\ \\ < =>x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=0\\ \\ < =>\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)
có vế trái luôn dương, vế phải = 0 => vô nghiệm
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
<=>\(\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+\dfrac{x-3}{2010}-1+...+\dfrac{x-2012}{1}-1=0\)
<=>\(\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)
<=>\(\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+1\right)=0\)
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...
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
\(\Leftrightarrow\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+...+\dfrac{x-2012}{1}-1=0\)
\(\Leftrightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+...+\dfrac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}\right)=0\)
Dễ thấy: \(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}>0\)
\(\Rightarrow x-2013=0\Rightarrow x=2013\)
Sao lại trừ 1 vậy bạn ??? mình không hiểu cho lắm mong bạn giúp đỡ
x2+y2+z2= xy+yz+zx.
=> 2x2+2y2+2z2-2xy-2yz-2zx=0
=> ( x-y)2+(y-z.)2+(z-x)2 =0
=> x=y=z=0
Thay x=y=z vào x2011+y2011+z2011=32012 ta được:
3.x2011=3.32011
=> x2011=32011
=> x=3 hoặc x = -3
Hay x=y=z=3 hoặc x=y=z=-3
1) có bn giải rồi ko giải nữa
2) \(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(2011^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(2012^4+\frac{1}{4}\right)}\)
Với mọi n thuộc N ta có :
\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2=\left(n^2-n+\frac{1}{2}\right)\left(n^2+n+\frac{1}{2}\right)\)
\(=\left[n\left(n-1\right)+\frac{1}{2}\right]\left[n\left(n+1\right)+\frac{1}{2}\right]\)
Áp dụng ta được :
\(A=\frac{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)....\left(2011.2012+\frac{1}{2}\right)}{\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right).......\left(2012.2013+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{2012.2013+\frac{1}{2}}=\frac{1}{8100313}\)
\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
\(\Leftrightarrow\dfrac{x-3}{2011}+\dfrac{x-2}{2012}-2=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}-2\)
\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-2}{2012}-1\right)=\left(\dfrac{x-2012}{2}-1\right)+\left(\dfrac{x-2011}{3}-1\right)\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=0\)
\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
<=>\(\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\)
<=>\(\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-2014}{3}\)
<=>\(\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
vì 1/2011+1/2012-1/2-1/3 khác 0
=>x-2014=0<=>x=2014
vậy....................
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+...+\dfrac{x-2012}{2}=2012\)
\(\Rightarrow\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+...+\dfrac{x-2012}{2}-2012=0\)
\(\Rightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1+...+\dfrac{x-2012}{2}-1=0\)
\(\Rightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}+...+\dfrac{x-2014}{2}=0\)
\(\Rightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{2}\right)=0\)
Mà \(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{2}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
\(x= \dfrac{2011^3-1}{2011^2+2012} = \dfrac{(2011-1)(2011^2+2011+1)}{2011^2 + 2011 + 1} = 2010\)
\(y = \dfrac{2012^3+1}{2012^2-2011} = \dfrac{(2012+1)(2012^2-2012+1)}{2012^2-2012 + 1} = 2013\)
Suy ra:
x + y = 2010 + 2013 = 4023