So sánh 2 số sau bằng cách vận dụng hằng đẳng thức:
A=4(32+1)(34+1).....(364+1) vs B=3128-1
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27 tháng 6 2023
`A=4(3^2+1)(3^4+1)...(3^64+1)`
`=>2A=(3^2-1)(3^2+1)(3^4+1)...(3^64+1)`
- Ta có:
`(3^2-1)(3^2+1)=3^4-1`
`(3^4-1)(3^4+1)=3^16-1`
`....`
`(3^64-1)(3^64+1)=3^128-1`
Suy ra `2A=3^128-1=B`
`=>A<B`
NL
Nguyễn Lê Phước Thịnh
CTVHS
15 tháng 6
a: \(A=1999\cdot2001\)
\(=\left(2000-1\right)\left(2000+1\right)\)
\(=2000^2-1=B-1\)
=>A<B
b: \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
=A-1
=>B<A
c: \(A=2011\cdot2013\)
\(=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)
=B-1
=>A<B
d: \(A=4\left(3^2+1\right)\left(3^4+1\right)\cdot\ldots\cdot\left(3^{64}+1\right)\)
\(=\frac12\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\ldots\cdot\left(3^{64}+1\right)\)
\(=\frac12\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\cdot\ldots\cdot\left(3^{64}+1\right)\)
\(=\frac12\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\cdot\left(3^{32}+1\right)\cdot\left(3^{64}+1\right)\)
\(=\frac12\left(3^{16}-1\right)\left(3^{16}+1\right)\cdot\left(3^{32}+1\right)\cdot\left(3^{64}+1\right)\)
\(=\frac12\left(3^{32}-1\right)\cdot\left(3^{32}+1\right)\cdot\left(3^{64}+1\right)\)
\(=\frac12\left(3^{64}-1\right)\cdot\left(3^{64}+1\right)=\frac12\left(3^{128}-1\right)\)
=1/2B
=>A<B
10 tháng 10 2018
\(A=4\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^{128}-1\right)< B\)
10 tháng 10 2018
\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1=B\)
\(\Rightarrow A< B\)
S
12 tháng 7 2018
a, \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1< 2000^2=B\)
Vậy A<B
b, \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=A\)
Vậy A>B
30 tháng 9 2019
\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{22}+1\right)\left(3^{64}+1\right)\)
\(2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(2A=3^{128}-1\Rightarrow A=\frac{3^{128}-1}{2}< 3^{128}-1=B\)
Vậy \(A< B\)
Chúc bạn học tốt !!!
VT
30 tháng 9 2019
A.(32-1)=4.(32-1)(32+1)(34+1)...(364+1)=4.(34-1)(34+1)...(364+1)= ... =4.(3128-1)
<=>8A=4B <=>2A=B =>B>A
AH
Akai Haruma
Giáo viên
15 tháng 8 2021
Lời giải:
a.
$27A=x^3-9x^2+162x-27=(x-3)^3+135x$
$=(303-3)^3+135.303=27040905$
$A=1001515$
b.
$B=2[(x+y)^3-3xy(x+y)]-3[(x+y)^2-2xy]$
$=2(1-3xy)-3(1-2xy)=2-6xy-3+6xy=-1$
c.
$C=x^3+y^3+3xy(x+y)=(x+y)^3=1^3=1$
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(.........\)
\(=\frac{1}{2}\left(3^{168}-1\right)\)\(< \)\(3^{168}-1\)
\(\Rightarrow\)\(A< B\)
Tại sao 4 lại trở thành 2 vậy. Giải thích giúp mình nhé.