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Ta có :
\(\frac{1}{2013}M=\frac{2013^{2012}+2012}{2013^{2012}+2013}=\frac{2013^{2012}+2013}{2013^{2012}+2013}-\frac{1}{2013^{2012}+2013}=1-\frac{1}{2013^{2012}+2013}\)
Lại có :
\(\frac{1}{2013}N=\frac{2013^{2011}+2012}{2013^{2011}+2013}=\frac{2013^{2011}+2013}{2013^{2011}+2013}-\frac{1}{2013^{2011}+2013}=1-\frac{1}{2013^{2011}+2013}\)
Vì \(\frac{1}{2013^{2012}+2013}< \frac{1}{2013^{2011}+2013}\) nên \(M=1-\frac{1}{2013^{2012}}>N=1-\frac{1}{2013^{2011}+2013}\)
Vậy \(M>N\)
Chúc bạn học tốt ~
Ta có \(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\)\(1+\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}=1+\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}\)
\(\Rightarrow\)\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
\(\Rightarrow\)\(\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)
\(\Rightarrow\)x+2014=0
\(\Rightarrow x=0-2014=-2014\)
Vậy x=-2014
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Leftrightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
\(\Leftrightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
\(\Leftrightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)
\(\Leftrightarrow x+2014=0\).Do \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
\(\Leftrightarrow x=-2014\)
Ta có :
x-y-z=0 => y+z=x (*(
Thay (*) và đa thức M ta có :
M=\(xyz-xy^2-xz^2=\left(y+z\right)yz-\left(y+z\right)y^2-\left(y+z\right)z^2\)
=\(y^2z+yz^2-y^3-zy^2-z^2y-z^3\)
=\(\left(y^2z-y^2z\right)-\left(z^2y-z^2y\right)-\left(y^3+z^3\right)\)
=\(-\left(y^3+z^3\right)\)
Mà \(-\left(y^3+z^3\right)\) là số đối của \(\left(y^3+z^3\right)\) nên M và N là 2 đa thức đối nhau.
Câu 1 :
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
=\(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.......+\dfrac{1}{2012}\right)\)=\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2013}\)=P
Vậy S=P
\(\frac{x+1}{2013}+\frac{x+2}{2012}=\frac{x+3}{2011}+\frac{x+4}{2010}\)
\(\Rightarrow\left(\frac{x+1}{2013}+1\right)+\left(\frac{x+2}{2012}+1\right)=\left(\frac{x+3}{2011}+1\right)+\left(\frac{x+4}{2010}+1\right)\)
\(\Rightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}-\frac{x+2014}{2011}-\frac{x+2014}{2010}=0\)
\(\Rightarrow\left(x+2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Vì \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)nên để \(\left(x+2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Thì x+2014=0
=>x=-2014
\(\frac{x+1}{2013}+\frac{x+2}{2012}=\frac{x+3}{2011}+\frac{x+4}{2010}\)
=> \(\frac{x+1+2013}{2013}+\frac{x+2+2012}{2012}=\frac{x+3+2011}{2011}+\frac{x+4+2010}{2010}\)
=> \(\frac{x+2014}{2013}+\frac{x+2014}{2012}=\frac{x+2014}{2011}+\frac{x+2014}{2010}\)
=> \(\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
=> \(x+2014=0\)(do \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\))
=> \(x=-2014\)
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
\(\frac{B}{A}=\frac{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}\)
\(=\frac{\left(\frac{2011}{2}+1\right)+\left(\frac{2010}{3}+1\right)+...+\left(\frac{1}{2012}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}\)
\(=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+....+\frac{2013}{2012}+\frac{2013}{2013}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}=2013\)