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Ta có: \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(\Rightarrow P-S=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2013}\right)=0\)
\(\Rightarrow\left(P-S\right)^{2013}=0^{2013}=0\)
Vậy \(\left(P-S\right)^{2013}=0\)
Ta có :
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+..........+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+..........+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+......+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+..........+\dfrac{1}{2013}\)
\(\Leftrightarrow S-P=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
\(1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-2\times\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2010}+\dfrac{1}{2012}\right)\)
\(\Rightarrow1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1005}+\dfrac{1}{1006}\right)\)
\(\Rightarrow\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(\Rightarrow S=P\Rightarrow S-P=0\Rightarrow\left(S-P\right)^{2013}=1\)
B=\(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)-2\(\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{2012}+\dfrac{1}{2013}\)=S
( A-B)2013 =0
Chúc ban học tốt nhé...!
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}=P\)
Vậy \(S=P\)
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
ta có:
\(P=\dfrac{1}{1007}+...+\dfrac{1}{2013}\\ \Rightarrow P=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}\right)\\ \Rightarrow P=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\\ \Rightarrow P=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\\ \Rightarrow P=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)\(\Rightarrow P=1-\dfrac{1}{2}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}=S\)
=> \(\left(P-S\right)^{2013}=0\)
vậy \(\left(P-S\right)^{2013}=0\)
1) So sánh các lũy thừa
a.
4444\(^{3333}\) và 3333\(^{4444}\)
4444\(^{3333}\) =(4\(^3\)\()\) \(^{111}\)
3333\(^{4444}\) =\((\)3\(^4\)\()\) \(^{111}\)
\(\rightarrow\) (4\(^3\)\()\) \(^{111}\) =64\(^{111}\) ; \((\)3\(^4\)\()\) \(^{111}\) =81\(^{111}\)
\(\rightarrow\)64\(^{111}\) <81\(^{111}\)
\(\Rightarrow\) 4444\(^{3333}\) < 3333\(^{4444}\)
Lười làm quá,ý còn lại bn làm tương tự,có ý lấy số chung để so sánh,có ý lấy số mũ để so sánh,có ý như trên.
\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}=\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\)
\(\Rightarrow\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1=\dfrac{x+2}{2013}+1+\dfrac{x+1}{2014}+1\)
\(\Rightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}-\dfrac{x+2015}{2013}-\dfrac{x+2015}{2014}=0\)
\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\ne0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
Vậy x = -2015
\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\)
\(\Rightarrow\dfrac{x+4}{2011}+\dfrac{x+3}{2012}-\dfrac{x+2}{2013}-\dfrac{x+1}{2014}=0\)
\(\Rightarrow\)\(\left(\dfrac{x+4}{2011}+1\right)+\left(\dfrac{x+3}{2012}+1\right)-\left(\dfrac{x+2}{2013}+1\right)-\left(\dfrac{x+1}{2014}+1\right)=0\)\(\Rightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}-\dfrac{x+2015}{2013}-\dfrac{x+2015}{2014}=0\)
\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\\ =\left(1+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}\right)\\ =\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\\ \Rightarrow S-P=0\\ \Rightarrow\left(S-P\right)^{2018}=0\)