Chứng minh rằng:
(2+1).(2²+1).(2⁴+1).(2^8+1).(2^16+1)=2³²-1
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\(\left(1-x\right)\left(1+x+x^2+...+x^{31}\right)=1-x^{32}\)
\(\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)
\(=\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)
\(=\left(1-x^4\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)
\(=\left(1-x^8\right)\left(1+x^8\right)\left(1+x^{16}\right)\)
\(=\left(1-x^{16}\right)\left(1+x^{16}\right)\)
\(=1-x^{32}\)
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 \(A< B\)
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
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
Chứng minh rằng:
a) 1/2-1/4+1/8-1/16+1/32-1/64<1/3
b) 1/3-2/3^2+3/3^3-3/3^4+...+99/3^99-100/3^100<3/16