So Sánh
A=(2+1).(2^2+1).(2^4+1).(2^8+1).(2^16+1) và B=2^32
So sánh A và B biết: A=(2+1)×(2^2+1)×(2^4+1)×(2^8+1)×(2^16+1)×(2^16+1) và B=2^32
Ta có: \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}\)
\(\Leftrightarrow A< B\)
SO SÁNH A = 3(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1 VÀ B =2^32
A = 3(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1
=(22-1)(22+1)(24+1)(28+1)(216+1)+1
=(24-1)(24+1)(28+1)(216+1)+1
=(28-1)(28+1)(216+1)+1
=(216-1)(216+1)+1
=232-1+1
=232 = B
vậy A=B
So sánh A và B biết: A=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1) và B=2^32
A = (2 - 1)(2 + 1)(2^2 + 1 )(2^4 + 1 ) (2^8 + 1)(2^16 + 1) ( nhân vói 2 - 1 = 1 Gía không thay dổi)
A = ( 2 ^2 - 1 )(2^2 + 1 )(2^4 + 1 )(2^8 + 1 )(2^16 + 1 )
A = ( 2^4 - 1 )(2^4 + 1)(2^8 + 1)(2^16 + 1)
A = (2^8 - 1)(2^8 + 1)(2^16 + 1)
A = (2^16 - 1)(2^16 + 1 )
A = 2^32 - 1 <2^32 = B
VẬy A < B
so sánh M = 2^32 và N = (2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)(2^16 + 1)
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
=>N<M
So sánh A = (2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1) và B = 2^32
Giúp tớ với mọi người ơi!!!!!
Ta có (21 -1)(21 + 1) = 22 - 1
(22 - 1)(22 + 1) = 24 - 1
tương tự như vậy ta sẽ có (2 -1)A = 232 - 1
vậy A < 232
CHO A=1+2+2^2+2^3+..+2^9;B=5.2^8.SO SÁNHA VÀ B
A=1+2+2^2+2^3+....+2^9
2A=2+2^2+2^3+....+2^10
2A-A=2^10-1
A=2^10-1/2
B=5.2^8=(2^2+1).2^8=2^10+2^8
=>B>A
2A = 2(1 + 2 + 22 + .... + 29 )
= 2 + 22 + 23 + ..... + 210
2A - A = (2 + 22 + 23 + ..... + 210) - (1 + 2 + 22 + .... + 29 )
A = 210 - 1
B = 5.28 = (22 + 1).28 = 210 + 28
210 - 1 < 210 + 28
=> A < B
So sánh 2 số A và B biết :
A = (3+1)(2^2+1)(3^4+1)(3^8+1)(3^16+1) và B = 3^32 - 1
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 = (2-1)(2+1)(2^2 + 1 ) (2^4 + 1 ) ( 2^8 + 1) ( 2^16 + 1)
A = (2^2 - 1)(2^2 + 1 ) ( 2^4 + 1 )(2^8 + 1 )(2^16 + 1)
A= ( 2^4 - 1 )( 2^4 + 1 )(2^8 + 1 )(2^16 + 1 )
A = (2^8 - 1 )(2^8 + 1 )(2^16 + 1 )
A = (2^16 - 1 )(2^16 + 1 )
A = 2^32 - 1 < 2^32 = B
Vậy A = B
k mik nka !
so sanh A va B
A= (2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
B=2^32
Ax(2-1)=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)=(2^4-1)(2^4+1)(2^8+1)(2^16+1)=(2^8-1)(2^8+1)(2^16+1)=(2^16-1)(2^16+1)=2^32-1
Vậy A=B
Áp dụng hằng đẵng thức A^2-B^2 đó bạn
Bài 6: So sánh
a,\(\dfrac{1}{2}\)+\(\dfrac{1}{_{ }2^2}\)+\(\dfrac{1}{2_{ }^3}\)+...+\(\dfrac{1}{2^{2014}}\)và 1 b,\(\dfrac{10^{2018}+5}{10^{2018}-8}\)và \(\dfrac{10^{2019}+5}{10^{2019}-8}\)
c,\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{23.24.25}\)và\(\dfrac{1}{4}\)