Đáp án + giải thích các bước giải:
`a, A = 2^0 + 2^1 + 2^2 + .... + 2^2020`
`2A = 2 . [2^0 + 2^1 + 2^2 + ... + 2^2020]`
`2A = 2^1 + 2^2 + 2^3 + .... + 2^2021`
` 2A - A = [2^1 + 2^2 + 2^3 + .... + 2^2021] - [2^0 + 2^1 + 2^2 + .... + 2^2020]`
` A = 2^1 + 2^2 + 2^3 + .... + 2^2021 - 1 - 2^1 - 2^2 - .... - 2^2020`
` A = 2^2021 - 1`
Mà `B = 2^2021 - 1` nên `A = B`
`b,` Ta có:
`A = 2009 . 2011`
` = [2010 - 1] . 2011`
` = 2010 . 2011 - 1 . 2011`
` = 2010 . 2011 - 2011`
`B = 2010^2`
`= 2010 . 2010`
` = [2011 - 1] . 2010`
` = 2011 . 2010 - 1 . 2010`
` = 2011 . 2010 - 2010`
Vì `2011 . 2010 > 0`
`=> 2011 . 2010 - 2010 > 2011 . 2010 - 2011`
`=> 2009 . 2011 < 2010^2`
`=> A < B`
`c,` Ta có:
`10^30 = [10^3]^10 = 1000^10`
`2^100 = [2^10]^10 = 1024^10`
Mà `1024 > 1000` nên `1024^10 > 1000^10`
Vậy `2^100 > 10^30`
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Tính chất :
`a . [b - 1] = a . b - a`
`[a^b]^c = a^[b . c]`
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\[A=2^0+2^1+2^2+2^3+-+2^{2010}\]
\[2A=2^1+2^2+2^3+2^4+-.+2^{2011}\]
\[2A-A=\left[2^1+2^2+2^3+2^4+-.+2^{2011}\right]-\left[2^0+2^1+2^2+2^3+-+2^{2010}\right]\]\[A=2^{2011}-2^0=2^{2011}-1\]
\[A=B=2^{2011}-1\]
\[A=2009.2011=2009.\left[2010+1\right]=2009.2010+2011\]
\[B=2010^2=2010.2010=2010\left[2009+1\right]=2010.2009+2010\]
\[A>B\]
\[A=10^{30}=\left[10^3\right]^{10}=1000^{10}\]
\[B=2^{100}=\left[2^{10}\right]^{10}=1024^{10}\]
\[A< B\]
\[A=333^{444}=\left[333^4\right]^{111}=\left[3^4.111^4\right]^{111}=\left[81.111^4\right]^{111}\]
\[B=444^{333}=\left[444^3\right]^{111}=\left[4^3.111^3\right]^{111}=\left[64.111^3\right]^{111}\]
\[A>B\]
\[A=3^{150}\]
\[B=5^{300}=\left[5^2\right]^{150}=25^{150}\]
\[A< B\]
Hay nhất
A = 2^0 +2^1 + 2^2 + 2^3 +...+2^2010 và B =2^2011 - 1
Ta có :
2A =2^1 +2^2 + 2^3 + 2^4 +...+2^2011
2A - A =[ 2^1 +2^2 + 2^3 + 2^4 +...+2^2011 ] -[ 1 + 2^2 + 2^3 +...+2^2010]
A =2^2011 - 1
MàB =2^2011 - 1
⇒ A = B
Bài viết tham gia Hoa điểm 10 2016
So sánh
a] 2^0 + 2^1 + 2^2 + 2^3 +...+2^2010 Và B = 2^2011 - 1
b] A = 2009 . 2011 và B = 2010^2