So sánh A=2^0+2^1+2+2+2^3 + 2^2010 và B=2 2011 1
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` ________________________________________________________ Tính chất : `a . (b - 1) = a . b - a` `(a^b)^c = a^(b . c)`
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\(A=2^0+2^1+2^2+2^3+-+2^{2010}\) \(2A=2^1+2^2+2^3+2^4+-.+2^{2011}\) \(2A-A=\left(2^1+2^2+2^3+2^4+-.+2^{2011}\right)-\left(2^0+2^1+2^2+2^3+-+2^{2010}\right)\)\(A=2^{2011}-2^0=2^{2011}-1\) \(A=B=2^{2011}-1\) \(A=2009.2011=2009.\left(2010+1\right)=2009.2010+2011\) \(B=2010^2=2010.2010=2010\left(2009+1\right)=2010.2009+2010\) \(A>B\) \(A=10^{30}=\left(10^3\right)^{10}=1000^{10}\) \(B=2^{100}=\left(2^{10}\right)^{10}=1024^{10}\) \(A< B\) \(A=333^{444}=\left(333^4\right)^{111}=\left(3^4.111^4\right)^{111}=\left(81.111^4\right)^{111}\) \(B=444^{333}=\left(444^3\right)^{111}=\left(4^3.111^3\right)^{111}=\left(64.111^3\right)^{111}\) \(A>B\) \(A=3^{150}\) \(B=5^{300}=\left(5^2\right)^{150}=25^{150}\) \(A< B\)
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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 |