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N
21 tháng 11 2018
\(\Leftrightarrow\frac{x^{2014}}{a^2+b^2+c^2+d^2}+\frac{y^{2014}}{a^2+b^2+c^2+d^2}+\frac{z^{2014}}{a^2+b^2+c^2+d^2}+\frac{t^{2014}}{a^2+b^2+c^2+d^2}\)
\(-\frac{x^{2014}}{a^2}-\frac{y^{2014}}{b^2}-\frac{z^{2014}}{c^2}-\frac{t^{2014}}{d^2}=0\)
\(\Leftrightarrow\left(\frac{x^{2014}}{a^2+b^2+c^2+d^2}-\frac{x^{2014}}{a^2}\right)+\left(\frac{y^{2014}}{a^2+b^2+c^2+d^2}-\frac{y^{2014}}{b^2}\right)+\left(\frac{z^{2014}}{a^2+b^2+c^2+d^2}-\frac{z^{2014}}{c^2}\right)\)
\(+\left(\frac{t^{2014}}{a^2+b^2+c^2+d^2}-\frac{t^{2014}}{d^2}\right)=0\)
\(\Leftrightarrow x^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+y^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\right)+\)
\(z^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\right)+t^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\right)=0\)
vì a2,b2,c2,d2 lớn hơn hoặc bằng 0
=> \(\hept{\begin{cases}\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\ne0\end{cases}}và....\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\ne0\)
\(\Rightarrow\hept{\begin{cases}x^{2014}=0\\y^{2014}=0\\z^{2014}=0\end{cases}}và..t^{2014}=0\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}và...t=0\)
=> \(\hept{\begin{cases}x^{2015}=0\\y^{2015}=0\\z^{2015}=0\end{cases}}và..t^{2015}=0\Rightarrow x^{2015}+y^{2015}+z^{2015}+t^{2015}=0\)
vậy \(x^{2015}+y^{2015}+z^{2015}+t^{2015}=0\)
N
2
TM
12 tháng 5 2015
2 S = 22016 - ( 22015 + 2 2014 + 22013 +.....+ 23 + 2 2 + 2 )
2S - S = 2 2016 + 1
13 tháng 5 2015
S = 22015- 22014- 22013-.......-22-21-20
2S = 22016 - 22015 -22014 - 22013 -..........- 23 -22 -21
2S -S = 22016 -22015 -22014 -22013 -....- 23-22 -21 - 22015 + 22014 + 22013 +.....+ 23 +22+21+20
= 22016 - 2x22015 + 20
= 20=1
1 tháng 8 2017
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(S=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Rightarrow\left(S-P\right)^{2016}=\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{1009}-...-\frac{1}{2015}\right)^{2016}=0^{2016}=0\)
1 tháng 8 2017
Ta thấy:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)+\frac{1}{2015}\)
\(S=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\) nên:
\(S=P\)\(\Rightarrow S-P=0\)\(\Rightarrow\left(S-P\right)^{2016}=0\)
N
0
HT
2
22 tháng 9 2017
Ta có :
\(S=2^{2015}-2^{2014}-..............-2-1\)
\(\Leftrightarrow S=2^{2015}-\left(2^{2014}+2^{2013}+...........+2+1\right)\)
Đặt :
\(A=2^{2014}+2^{2013}+.........+2+1\)
\(\Leftrightarrow2A=2^{2015}+2^{2014}+.............+2\)
\(\Leftrightarrow2A-A=\left(2^{2015}+2^{2014}+..........+2\right)-\left(2^{2014}+2^{2013}+.........+1\right)\)
\(\Leftrightarrow A=2^{2015}-1\)
\(\Leftrightarrow S=2^{2015}-\left(2^{2015}+1\right)\)
\(\Leftrightarrow S=2^{2015}-2^{2015}+1\)
\(\Leftrightarrow S=0+1=1\)
22 tháng 9 2017
\(S=2^{2015}-2^{2014}-2^{2013}-...2-1\)
\(2S=2^{2015}-2^{2014}-2^{2013}-...-2\)
\(2S-S=2^{2015}-2^{2014}-2^{2014}-2^{2013}+2^{2013}-...-2+2+1\)
\(S=2^{2015}-2.2^{2014}+1\)
\(S=2^{2015}-2^{2015}+1=1\)
Tham khảo, chúc bạn học giỏi! Haizzz
CS
29 tháng 3 2018
Ta có:
*) \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}\)
\(\Rightarrow S=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2014}\right)\)
\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1007}\right)\)
\(\Rightarrow S=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
Vậy \(\left(S-B\right)^{2016}=\left[\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)\right]^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0\)
19 tháng 9 2016
a) Đặt \(A=2^{2016}-2^{2015}+2^{2014}-2^{2013}+...+2^2-2^1\)
\(\Rightarrow2A=2^{2017}-2^{2016}+2^{2015}-2^{2014}+...+2^3-2^2\)
\(\Rightarrow2A+A=\left(2^{2017}-2^{2015}+2^{2014}-2^{2013}+...+2^3-2^2\right)+\left(2^{2016}-2^{2015}+2^{2014}-2^{2013}+...+2^2+2^1\right)\)
\(\Rightarrow3A=2^{2017}+1\)
\(\Rightarrow A=\frac{2^{2017}+1}{3}\)
b) Đặt \(B=3^{1000}-3^{999}+3^{998}-3^{997}+...+3^2-3^1+3^0\)
\(\Rightarrow3B=3^{1001}-3^{1000}+3^{999}-3^{997}+...+3^3-3^2+3^1\)
\(\Rightarrow3B+B=\left(3^{1001}-3^{1000}+3^{999}-3^{998}+...+3^3-3^2+3^1\right)+\left(3^{1000}-3^{999}+3^{998}-3^{997}+...+3^2-3^1+3^0\right)\)
\(\Rightarrow4B=3^{1001}+3^0\)
\(\Rightarrow B=\frac{3^{1001}+1}{4}\)
19 tháng 9 2016
a) Đặt A = 22016 - 22015 + 22014 - 22013 + ... + 22 - 21
2A = 22017 - 22016 + 22015 - 22014 + ... + 23 - 22
2A + A = (22017 - 22016 + 22015 - 22014 + ... + 23 - 22) + (22016 - 22015 + 22014 - 22013 + ... + 22 - 21)
3A = 22017 - 21
3A = 22017 - 2
\(A=\frac{2^{2017}-2}{3}\)
b) lm tương tự câu a