ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều 

Set \(S=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2023}}\)

Then \(3S=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2022}}\)

Hence \(2S=3S-S=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2022}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2023}}\right)\)

\(=1-\dfrac{1}{3^{2023}}\)

\(\Leftrightarrow S=\dfrac{1}{2}-\dfrac{1}{2.3^{2023}}< \dfrac{1}{2}\) (Q. E. D)

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)

Ta có: \(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)

\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)

\(2A=1-\dfrac{1}{3^{2023}}\)

\(A=\dfrac{1-\dfrac{1}{3^{2023}}}{2}\)

Vì \(\dfrac{1-\dfrac{1}{3^{2023}}}{2}< \dfrac{1}{2}\) nên \(A< \dfrac{1}{2}\)

Vậy...

A=1/3+1/3^2+...+1/3^2005

=> 3A= 1+1/3+...+1/3^2004

=> 3A-A=(1+1/3+...+1/3^2004)-(1/3+1/3^2+...+1/3^2005)

=> 2A =1-1/3^2005 <1 

=> A<1/2

1,2 : 10 = 0,12
4,6 : 1000 = 0,0046
781,5 : 100 = 7,815
15,4 : 100 = 0,154
45,82 : 10 = 4,582
15632 : 1000 = 15,632
hok tốt nha ^_^

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)

\(\frac{A}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}.....+\frac{1}{3^{100}}+\frac{1}{3^{101}}\)

\(A-\frac{A}{3}=\frac{2A}{3}=\frac{1}{3}=\frac{1}{3}-\frac{1}{3^{101}}\Rightarrow2A=1-\frac{1}{3^{100}}\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{100}}< \frac{1}{2}\)

\(\frac{1}{3}M=\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{100}}\)

\(M-\frac{1}{3}M=\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+....+\left(\frac{1}{3^{99}}-\frac{1}{3^{99}}\right)+\frac{1}{3}-\frac{1}{3^{100}}\)

\(\frac{2}{3}M=\frac{1}{3}-\frac{1}{3^{100}}\)

Vậy \(M=\left(\frac{1}{3}-\frac{1}{3^{100}}\right):\frac{2}{3}=\frac{1}{2}-\frac{1}{2.3^{99}}<\frac{1}{2}\)

KL: M    < 1/2 (dpcm)

Here........ PDZ