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31 tháng 1 2018
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2015}-1\right)\left(\dfrac{1}{2016}-1\right)\left(\dfrac{1}{2017}-1\right)\\ A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)...\left(-\dfrac{2014}{2015}\right)\left(-\dfrac{2015}{2016}\right)\left(-\dfrac{2016}{2017}\right)\\ A=\dfrac{1.2.3.4...2014.2015.2016}{2.3.4...2015.2016.2017}=\dfrac{1}{2017}\)
\(B=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{2015}\right)\left(-1\dfrac{1}{2016}\right)\left(-1\dfrac{1}{2017}\right)\\ B=\left(-\dfrac{3}{2}\right)\left(-\dfrac{4}{3}\right)\left(-\dfrac{5}{4}\right)...\left(-\dfrac{2016}{2015}\right)\left(-\dfrac{2017}{2016}\right)\left(-\dfrac{2018}{2017}\right)\\ B=\dfrac{3.4.5...2016.2017.2018}{2.3.4...2015.2016.2017}=\dfrac{2018}{2}=1009\)
\(M=A.B=\dfrac{1}{2017}.1009=\dfrac{1009}{2017}\)
KR
1
NT
13 tháng 10 2017
\(A=2^{2017}-\left(2^{2016}+...+2^1+1\right)\\ \)
Đặt \(B=1+2+...+2^{2016}\)
\(\Rightarrow2.B=2+2^2+...+2^{2017}\\ \Rightarrow2.B-B=\left(2+2^2+...+2^{2017}\right)-\left(1+2+...+2^{2016}\right)\\ \Rightarrow B=2^{2017}-1\\ \Rightarrow A=2^{2017}-B=2^{2017}-2^{2017}+1=1\)
PT
1
19 tháng 11 2016
3 + |x - 3|2016 = 22017 - 22016 - 22016 - ... - 22
3 + |x - 3|2016 = 22017 - (22016 + 22015 + ... + 22)
Đặt A = 22016 + 22015 + ... + 22
2A = 22017 + 22016 + ... + 23
2A - A = 22017 - 22
A = 22017 - 4
3 + |x - 3|2016 = 22017 - (22017 - 4) = 22017 - 22017 + 4 = 4
=> |x - 3|2016 = 4 - 3 = 1
=> |x - 3| = 1
\(\Rightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
30 tháng 8 2016
\(\frac{10^{2016}+2^3}{9}=\frac{10^{2016}-1}{9}+\frac{2^3+1}{9}=\left(1+10+10^2+...+10^{2015}\right)+1\in N.\)
TN
0
NT
1
4 tháng 3 2019
\(P\left(x\right)=x^{2017}-2018x^{2017}+2018x^{2016}-...-2018x+1\)
Vì \(x=2017\)
\(\Leftrightarrow x+1=2018\)
Thay vào P(x) ta được :
\(P\left(x\right)=x^{2017}-x^{2017}\left(x+1\right)+x^{2016}\left(x+1\right)-...-x\left(x+1\right)+1\)
\(P\left(x\right)=x^{2017}-x^{2018}-x^{2017}+x^{2017}+x^{2016}-...-x^2-x+1\)
\(P\left(x\right)=-x^{2018}+1\)
\(P\left(x\right)=-2017^{2018}+1\)
Ta có :
\(M=2^{2017}-\left(2^{2016}+2^{2017}+...............+2+1\right)\)
Đặt :
\(A=2^{2016}+2^{2015}+................+2+1\)
\(\Leftrightarrow2A=2^{2017}+2^{2016}+2^{2015}+............+2^2+2\)
\(\Leftrightarrow2A-A=\left(2^{2017}+2^{2016}+........+2\right)-\left(2^{2016}+2^{2015}+..........+1\right)\)
\(\Leftrightarrow A=2^{2017}-1\)
\(\Leftrightarrow M=2^{2017}-A\)
\(\Leftrightarrow M=2^{2017}-\left(2^{2017}-1\right)\)
\(\Leftrightarrow M=2^{2017}-2^{2017}+1\)
\(\Leftrightarrow M=0+1=1\)
\(M=2^{2017}-\left(2^{2016}+2^{2015}+...+2^1+2^0\right)\)
Đặt :
\(S=2^{2016}+2^{2015}+...+2^1+2^0\)
\(\Rightarrow S=2^0+2^1+...+2^{2015}+2^{2016}\)
\(\Rightarrow2S=2\left(2^0+2^1+...+2^{2015}+2^{2016}\right)\)
\(\Rightarrow2S=2^1+2^2+...+2^{2016}+2^{2017}\)
\(\Rightarrow2S-S=\left(2^1+2^2+...+2^{2016}+2^{2017}\right)-\left(2^0+2^1+...+2^{2015}+2^{2016}\right)\)
\(\Rightarrow S=2^{2017}-1\)
Thay S vào M ta có:
\(M=2^{2017}-\left(2^{2017}-1\right)\)
\(M=2^{2017}-2^{2017}+1=1\)