a: \(A=2011\cdot2013=\left(2012-1\right)\left(2012+1\right)\)
\(=2012^2-1=B-1\)
=>A<B
b: \(A=4\left(3^2+1\right)\left(3^4+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\cdot...\cdot\left(3^{64}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\cdot\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(=\dfrac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^{128}-1\right)=\dfrac{1}{2}B\)
=>A<B
`a)A=2011.2013`
`=(2012-1)(2012+1)`
`=2012^2-1<2012^2=B`
`b)A=4(3^2+1)(3^4+1)...(3^64+1)`
`A=(3+1)(3^2+1)(3^4+1)...(3^64+1)`
`2A=(3-1)(3+1)(3^2+1)(3^4+1)...(3^64+1)`
`2A=(3^2-1)(3^2+1)(3^4+1)...(3^64+1)`
...
`2A=3^128-1=B`
`=>A<B`
\(a,A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)
\(=>A=2012^2-1< 2012^2\)
\(=>A< B\)
\(b,A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=>2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(=>2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=>2A=\left(3^8-1\right)...\left(3^{64}+1\right)\)
\(=>2A=3^{128}-1\)
\(=>A=\dfrac{3^{128}-1}{2}< 3^{128}-1\)
\(=>A< B\)
`#3107.101107`
`4.`
`a)`
`A = 2011 * 2013` và `B = 2012^2`
Ta có:
`2011 = 2012 - 1; 2013 = 2012 + 1`
`=> A = (2012 - 1)(2012 + 1) = 2012^2 - 1`
`2012^2 - 1 < 2012^2`
`=> A < B`
`b)`
`A = 4(3^2 + 1)(3^4 + 1)...(3^64 + 1)` và `B = 3^(128) - 1
Ta có:
`A = 4(3^2 + 1)(3^4 + 1)...(3^64 + 1)`
`2A = 8(3^2 + 1)(3^4 + 1)...(3^64 + 1)`
`2A = (3^2 - 1)(3^2 + 1)(3^4 + 1)...(3^64 + 1)`
`2A = (3^4 - 1)(3^4 + 1)....(3^64 + 1)`
`2A = (3^16 - 1)(3^16 + 1)...(3^64 + 1)`
`2A = (3^32 - 1)(3^32 + 1)(3^64 + 1)`
`2A = (3^64 - 1)(3^64 + 1)`
`2A = 3^(128) - 1`
`=> A = (3^(128)-1)/2 < 3^(128) - 1`
`=> A < B.`
