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![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{49}+\left(\frac{1}{5}\right)^{50}\)
\(5M=1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{48}+\left(\frac{1}{5}\right)^{49}\)
5M - M = \(1-\left(\frac{1}{5}\right)^{50}\)hay 4M = \(1-\left(\frac{1}{5}\right)^{50}\)< 1
\(\Rightarrow M=\frac{1-\left(\frac{1}{5}\right)^{50}}{4}< \frac{1}{4}\)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)(1)
\(\Rightarrow5M=1+\frac{1}{5}+...+\left(\frac{1}{5}\right)^{49}\)(2)
Lấy (2)-(1) ta có
\(\Rightarrow4M=1-\left(\frac{1}{5}\right)^{50}\)
\(\Rightarrow M=\frac{1-\frac{1}{5^{50}}}{4}\)
Do \(1-\frac{1}{5^{50}}< 1\)
\(\Rightarrow M< \frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 5M = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}\)
Lấy 5M trừ M ta có :
\(5M-M=\left(1+\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{2013}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+..+\frac{1}{5^{2014}}\right)\)
\(4M=1-\frac{1}{5^{2014}}\)
\(M=\left(1-\frac{1}{5^{2014}}\right):4=\frac{1}{4}-\frac{1}{5^{2014}.4}< \frac{1}{3}\)
\(\Rightarrow M< \frac{1}{3}\left(\text{ĐPCM}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
A=0,5-|x-4|
Vì -|x-4|\(\le\)0
Suy ra:0,5-|x-4|\(\le\)0,5
Dấu = xảy ra khi x-4=0;x=4
Vậy Max A=0,5 khi x=4
B=1,25+|5-x|
Vì |5-x|\(\ge\)0
Suy ra:1,25+|5-x|\(\ge\)1,25
Dấu = xảy ra khi 5-x=0;x=5
Vậy Min B=1,25 khi x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)
\(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\)
\(5M=5\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\right)\)
\(5M=1+\frac{1}{5}+...+\frac{1}{5^{49}}\)
\(5M-M=\left(1+\frac{1}{5}+...+\frac{1}{5^{49}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\right)\)
\(4M=1-\frac{1}{5^{50}}\)
\(M=\frac{1-\frac{1}{5^{50}}}{4}< \frac{1}{4}=0,25\)
Đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(A=-\left(x-2\right)^2+3\le3\)
Dấu '=' xảy ra khi x=2
b: \(B=-\left(2x-3\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=3/2
c: \(C=-\left(2x-\dfrac{3}{5}\right)^2-\dfrac{1}{3}\le-\dfrac{1}{3}\)
Dấu '=' xảy ra khi x=3/10
d: \(D=-\left|\dfrac{5}{3}-x\right|\le0\)
Dấu '=' xảy ra khi x=5/3
e: \(E=-\left|x-\dfrac{1}{10}\right|+9\le9\)
Dấu '=' xảy ra khi x=1/10
f: \(F=-\left|\dfrac{1}{2}x+1\right|-\dfrac{2}{3}\le-\dfrac{2}{3}\)
Dấu '=' xảy ra khi x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{25.5^3.1}{625.5^2}=\frac{5^2.5^3}{5^4.5^2}=\frac{1}{5}\)
các câu sau mk k hiểu bn viết j
=>5M=1+1/5+1/5^2+...+1/5^48+1/5^49
=>5M-M=(1+1/5+1/5^2+..+1/5^48+1/5^49)-(1/5+1/5^2+1/5^3+...+1/5^49+1/5^50)
=>4M=1-1/5^50
=>M=(1-1/5^50)/4
mà 1-1/5^50<1
=>M<1/4(đpcm)
tich nha ban