\(\left(1-x\right)\left(1+x+x^2+...+x^{31}\right)=1-x^{32}\)

\(\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)

\(=\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)

\(=\left(1-x^4\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)

\(=\left(1-x^8\right)\left(1+x^8\right)\left(1+x^{16}\right)\)

\(=\left(1-x^{16}\right)\left(1+x^{16}\right)\)

\(=1-x^{32}\)

Ta có đpcm. 

A=\(\frac{10^8+2}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=1+\frac{3}{10^8-3}\)

Vì\(10^8-1>10^8-3\)

\(\Rightarrow\frac{3}{10^8-1}< \frac{3}{10^8-3}\)

\(\Rightarrow1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\)

Vậy \(A< B\)

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bài này lần trước mik vừa trả lời lần trước

 Đặt A=1/2−1/4+1/8−1/16+1/32−1/64A

         =1/2−1/4+1/8−1/16+1/32−1/64
     2A=1−1/2+1/4−1/8+1/16−1/32

    2A =1−1/2+1/4−1/8+1/16−1/32
    3A=2A+A=1−1/64<1

   ⇒A<1/3

k cho minh nha

mình cũng khó câu này ai giúp mình với

Đặt \(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}.\)

\(\Rightarrow2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)

\(\Rightarrow2A+A=\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)\)

\(\Rightarrow3A=1-\frac{1}{64}< 1\)\(\Rightarrow A< \frac{1}{3}\left(đpcm\right)\)

                                  Vậy \(A< \frac{1}{3}.\)

 đặt A=1/2-1/4+1/8-1/16+1/32-1/64

2A=1-1/2+1/4-1/8+1/16-1/32

2A-A=1-1/64 A=63/64

Vì 63/64<1/3

nên 1/2-1/4+1/8-1/16+1/32-1/64<1/3

Vậy 1/2-1/4+1/8-1/16+1/32-1/64<1/3