SS:
A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) và B=3^32-1
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Có: \(A=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=...........................\)
\(=\frac{3^{32}-1}{2}\)
\(B=3^{32-1}\)
=> \(A< B\)
a) A = 2016.2018 = ( 2017 - 1 ).2018 = 2017.2018 - 2018 ( 1 )
B = 20172 = 2017.2017 = 2017.( 2018 - 1) = 2017.2018 - 2017 ( 2 )
Từ (1) và (2), ta thấy: - 2018 < - 2017 => 2017.2018 - 2018 < 2017.2018 - 2017 <=> 2016.2018 < 20172
Vậy A < B
~ Phần b khi nào nghĩ ra tớ sẽ làm ngay ạ :) Còn phần này chắc chắn đúng cậu nhé ~
b)\(x=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^{16}-1\right)\left(3^{16}+1\right)\Rightarrow x=\frac{3^{32}-1}{2}\)
Thấy \(x=\frac{3^{32-1}}{2}< 3^{32}-1=y\)
Mình ghi nhầm đề bài 1 tí đề bài là :
So sánh 2 số A và B biết :
A = (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) và B = 3^32 - 1
\(A=4.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=\frac{3^{32}-1}{2}< 3^{32}-1=B\)
Vậy \(A< B\)
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
TL:
Đặt A=(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)A=(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)
⇒2A=(3−1)(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)⇒2A=(3−1)(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)
=(32−1)(32+1)(34+1)(38+1)(316+1)(332+1)=(32−1)(32+1)(34+1)(38+1)(316+1)(332+1)
=...........................................=...........................................
=(332−1)(332+1)=364−1=(332−1)(332+1)=364−1
⇒A=364−122
k cho mk nha
HT