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Ta có:
B=2012^2.
=>B=2012*2012.
=>B=2012*2011+2012.
=>B=2011*2012+2011+1.
=>B=2011*(2012+1)+1.
=>B=2011*2013+1.
Mà A=2011*2013.
Vậy A<B.
Ta có:
\(A=2011\cdot2013=\left(2012-1\right)\left(2012+1\right)\)
\(=2012^2-1< 2012^2=B\)
VẬY A<B
\(A=1999.2000+1999\\ B=2000.1999+2000\)
Vì \(1999.2000+1999< 1999.2000+2000\)
\(=>A< B\)
Đúng thì tích nha :D
`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`
a)A=\(1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1\)
Vậy A < B
b) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=A\)
Vậy B < A
a) Ta có: \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)\)
\(=2000^2-1^2< 2000^2\)
Vậy A < B.
b) 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.
\(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1\)
\(< 2016^2=B\)
Nên A<B
\(B=2016^2\)
\(\Rightarrow B=\left(2017-1\right)^2\)
\(\Rightarrow B=2017^2-4034+1=2017^2-4033\)(1)
Lại Có :
\(A=2015.2017=\left(2017-2\right).2017\)
\(\Rightarrow A=2017^2-4034\)(2)
Từ (1) và (2) => B>A
\(A=4\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^{128}-1\right)< B\)
\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1=B\)
\(\Rightarrow A< B\)
Ta có : A = 1999 x 2001 = 1999 x (1 + 2000) = 1999 x 2000 + 1999
B = 2000 x 2000 = 2000 x (1999 + 1) = 2000 x 1999 + 2000
Vậy A < B
Sorry mk chưa đoc kĩ đề mk làm lại nhá
Áp dụng hàng đẳng thức (a - b)(a + b) = a2 - b2
Ta có : A = (2000 - 1)(2000 + 1) = 20002 - 1
Mà B = 20002
Nên A < B
Áp dụng hàng đẳng thức (a - b)(a + b) = a2 - b2
Ta có : A = (2012 - 1)(2012 + 1) = 20122 - 1
Mà B = 20122
Nên A < B