Nhầm

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

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

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

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

bạn thiếu ĐPCM

a<1\2

\(\Rightarrow\left(x-1\right)^2-\left(2x-3\right)^2=0\\ \Rightarrow\left(x-1-2x+3\right)\left(x-1+2x-3\right)=0\\ \Rightarrow\left(2-x\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)

Nguyễn Hoàng Minh  thank you

ta có: 2B=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)

B=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+..+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)

=>2B-B=\(1-\frac{1}{2^{99}}\)

mà 1/2^99>0 nên B<1 (đpcm)

3A= 1+ 1/3 + 1/3^2 + ... + 1/3^98

3A-A=1 + 1/3 + 1/3^2 + ... + 1/3^98 - 1/3 - 1/3^2 - 1/3^3 - .... - 1/3^99

2A= 1 - 1/3^99 < 1

=> A < 1/2

3A = 1+1/3+1/3^2+...+1/3^99

3A-A=(1+1/3+...+1/3^99)-(1/3+1/3^2+...+1/3^99)

2A= 1-1/3^99

A  = (1-1/3^99)/2 < 1/2

=> A < 1/2

ta có 

\(B=1+\left(1-\frac{1}{2}\right)+..+\left(1-\frac{1}{100}\right)\)

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

Vậy A=B