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\(\frac{10^{2016}+2^3}{9}=\frac{10^{2016}-1}{9}+\frac{2^3+1}{9}=\left(1+10+10^2+...+10^{2015}\right)+1\in N.\)
\(A=2^{2017}-\left(2^{2016}+...+2^1+1\right)\\ \)
Đặt \(B=1+2+...+2^{2016}\)
\(\Rightarrow2.B=2+2^2+...+2^{2017}\\ \Rightarrow2.B-B=\left(2+2^2+...+2^{2017}\right)-\left(1+2+...+2^{2016}\right)\\ \Rightarrow B=2^{2017}-1\\ \Rightarrow A=2^{2017}-B=2^{2017}-2^{2017}+1=1\)
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
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
\(\Rightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
a)Vì |x−2015|= 1/2 nên x-2015=-1/2 hoặc x-2015=1/2
Nếu x-2015=-1/2 thì
x=2015+(-1)/2
x=4029/2
Nếu x-2015=1/2 thì
x=2015+1/2
x=4031/2
Vậy x=4029/2
hoặc x=4031/2
b)
Nếu x>2016 thì |x−2015|=x-2015 ,|x−2016|=x-2016
Khi đó: |x−2015|+|x−2016|=2017
=>x-2015+x-2016=2017
=>2x-4031=2017
=>2x=6048=>x=3024(thỏa mãn x>2016)
Nếu 2015<x<2016 thì |x−2015|=x-2015,
|x−2016|=2016-x. khi đó
|x−2015|+|x−2016|=2017
=>x-2015+2016-x=2017
=>1=2017(vô lý loại)
Nếu x>2015 thì |x−2015|=2015-x,|x−2016|=2016-x
Khi đó:
|x−2015|+|x−2016|=2017
=>2015-x+2016-x=2017
=>4031-2x=2017
=>2x=2014=>x=1007(thỏa mãn x<2015)
Vậy x=1007 hoặc x=3024
a: \(\left(x-2\right)^2+\left(x-y\right)^6+3\ge3\)
\(\Leftrightarrow A=\dfrac{2003}{\left(x-2\right)^2+\left(x-y\right)^6+3}\le\dfrac{2003}{3}\)
Dấu '=' xảy ra khi x=y=2
b: \(B=-\left(2x+\dfrac{1}{3}\right)^6+3\le3\forall x\)
Dấu '=' xảy ra khi x=-1/6
c: \(C=\dfrac{x^{2016}+2015+2}{x^{2016}+2015}=1+\dfrac{2}{x^{2016}+2015}\le\dfrac{2}{2015}+1=\dfrac{2017}{2015}\)
Dấu '=' xảy ra khi x=0
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:
\(M=2^{2017}-\left(2^{2017}-1\right)\)
\(M=2^{2017}-2^{2017}+1=1\)