so sanh :
A = 4*(3^2+1)*(3^4+1)*(3^8+1)*...*(3^64+1)
B=3^128-1
So sánh :
a) A = 2^16 và B = (2+1).(2^2+1).(2^4+1).(2^8+1)
b) A = 4.(3^2+1).(3^4+1)....(3^64+1) và B = 3^128-1
Câu a : Ta có :
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1< 2^{16}\)
Vậy \(A>B\)
Câu b : Ta có :
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\dfrac{8\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)}{2}\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)}{2}\)
\(=\dfrac{...\left(3^{64}-1\right)\left(3^{64}+1\right)}{2}\)
\(=\dfrac{3^{128}-1}{2}< 3^{128}-1\)
Vậy \(A< B\)
So sánh 2 số bằng cách vận dụng hàng đẳng thức
a)A=2^16 và B=( 2+1)(2^2+1)(2^4+1)(2^8+1)
b)A=4(3^2+1)(3^4+1)...(3^64+1)và B=3^128 -1
So sánh:
A = 8 . (32+1). (34+1). ... .(364+1) với B = 3128-1
Xét biểu thức A
\(A=8\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(...=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1\)
Vậy \(A=B\)
So sánh: 4.(32+1).(34+1).(38+1)........(364+1) và 3128-1
\(S=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(\left(3^2-1\right)S=4\cdot\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(8S=4\cdot\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(2S=\left(3^8-1\right)\left(3^8+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
...
\(2S=3^{128}-1\)
Vậy S < 3128 - 1
Giúp mình vs ạ mai mình học rùi
So sánh 2 số sau bằng cách vận dụng hằng đẳng thức :
a) A = 1999.2001 và B = 20002
b) A = 2^16 và B = (2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)
c) A = 2011.2013 và B = 2012^2
d) A = 4(3^2 + 1)(3^4 + 1)....(3^64 + 1) và B = 3^128 - 1
Bài1:So sánh
A=2006/987654321+2007/246813579
B=2007/987654321+2006/246813579
b)1965/1976 và 1973/1975
Bài2:Tìm x
a)3 - (5 và 3/8 + x - 7 và 5/24):6 và 2/3=2
b) (1/1*2+1/2*3+1/3*4+1/4*5+1/5*6)*10 - x=0
Bài3:Tính nhanh
A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
1. 2006/987654321 + 2007/246813579 = 2007/246813579 + 2006/987654321
=>
2.
3 - (5.3/8 + X - 7 . 5/24) : 6 . 2/3 =2
3 - (15/8 + X - 35/24) : 4 = 2
3 - (15/8 + X - 35/24) = 2 . 4
3 - (15/8 + X - 35/24) = 8
15/8 + X - 35/24 = 3 - 8
15/8 + X - 35/24 = -5
15/8 + X = -5 + 35/24
15/8 + X = -85/24
X = -85/24 - 15/8
X = -65/12
thu gọn:
a) (2+1)(2^2+1)(2^4+1)..............(2^32+1)-2^64
b) (5+3)(5^2+3^2)(5^4+3^4)...................(5^64+3^64).\(\frac{5^{128}-3^{128}}{2}\)
So sánh 2 số sau bằng cách vận dụng hằng đẳng thức:
A=4(32+1)(34+1).....(364+1) vs B=3128-1
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(.........\)
\(=\frac{1}{2}\left(3^{168}-1\right)\)\(< \)\(3^{168}-1\)
\(\Rightarrow\)\(A< B\)
Tại sao 4 lại trở thành 2 vậy. Giải thích giúp mình nhé.
tính nhanh p/s 1+ 5/4 + 5/8 + 5/16 + 5/32 + 5/64
b) 1/3 +1/9 + 1/27 + 1/81 +...........+ 1/59049
c) 3/2 + 3/8 + 3/32 +3/128 + 3/512
d) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 giúp mình với
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