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bài 4 : c1 \(3^{4000}\)và \(9^{2000}\)
\(\Leftrightarrow9^{2000}\Leftrightarrow\left(3^2\right)^2^{000}\Leftrightarrow3^{4000}\)
vì \(3^{4000}=3^{4000}\Leftrightarrow3^{4000}=9^{2000}\)
c2
ta có
\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)
\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)
vì \(81^{1000}=81^{1000}\Leftrightarrow3^{4000}=9^{2000}\)
bài 5
\(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
vì \(8^{111}< 9^{111}\Leftrightarrow2^{332}< 3^{223}\)
3) M = 22010 - (22009 + 22008 + .... + 21 + 20)
Đặt N = 22009 + 22008 + .... + 21 + 20
=> 2N = 22010 + 22009 + .... + 22 + 21
=> 2N - N = (22010 + 22009 + .... + 22 + 21) - (22009 + 22008 + .... + 21 + 20)
=> N = 22010 - 1
Khi đó M = 22010 - (22010 - 1) = 1
4) C1 Ta có 34000 = (34)1000 = 811000 = (92)1000 = 92000
34000 = 92000
C2 Ta có : 34000 = (34)1000 = 811000 (1)
Lại có 92000 = (92)1000 = 811000 (2)
Từ (1) (2) => 34000 = 92000
5 Ta có 2332 < 2333 = (23)111 = 8111 < 9111 = (32)111 = 3222 < 3223
=> 2332 < 3223
2) Ta có n150 < 5225
=> (n5)75 < (53)75
=> n5 < 53
=> n5 < 125
Vì n là số nguyên lớn nhất => n = 2
\(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)
\(2^{2009}+2^{2008}+.......+2+1=b\)
\(\Rightarrow2b=2^{2010}+2^{2009}+.........+2^2+2\)
\(\Rightarrow2b-b=2^{2010}-1\Rightarrow b=2^{2010}-1\)
\(\Rightarrow A=2^{2010}-b=2^{2010}-\left(2^{2010}-1\right)=1\)
Đặt \(A=2^{2009}+2^{2008}+...+2^1+2^0\)
Ta có : \(2A=2^{2010}+2^{2009}+...+2^2+2^1\)
\(\Rightarrow2A-A=2^{2010}-2^0\Rightarrow A=2^{2010}-1\)
Do đó : \(M=2^{2010}-A=2^{2010}-\left[2^{2010}-1\right]=1\)
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(2^{2010}-M=2^{2009}+2^{2008}+...+2+1\)
\(2\left(2^{2010}-M\right)=2\left(2^{2009}+2^{2008}+...+2+1\right)\)
\(2\left(2^{2010}-M\right)=2^{2010}+2^{2009}+...+2^2+2\)
\(2\left(2^{2010}-M\right)-M=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)
\(2^{2010}-M=2^{2010}+2^{2009}+...+4+2-2^{2009}-2^{2008}-...-2-1\)
\(2^{2010}-M=2^{2010}-1\)
=> M = 1
\(M=2^{2010}-2^{2009}-2^{2008}-...-2^1-2^0\)
\(-M=-\left(2^{2010}-2^{2009}-2^{2008}-...-2^1-2^0\right)\)
\(-M=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)
\(-2M=2.\left(2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(-2M=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)
\(-M=2^{2011}+2^{2010}+...+2^2+2^1-\left(2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(-M=2^{2011}-1=>M=-2^{2011}+1\)
2332 < 2333 = (23)111 = 8111 hay 2332 < 8111
3223 > 3222 = (32)111 = 9111 hay 3223 > 9111
Mà 8111 < 9111
=> 2332 < 8111 < 9111 < 3223
Vậy 2332 < 3223.
Ta có:
2332 < 2333 = (23)111 = 8111
3223 > 3222 = (32)111 = 9111
Vì 8111 < 9111
=> 2332 < 3223
Ủng hộ mk nha ★_★^_-
Ta có: 2332 < 2333 = (23)111 = 8111
3223 > 3222 = (32)111 = 9111
Vì 8 < 9 Nên 8111 < 9111
Vậy 2332 < 3223
Đặt \(A=2^{2009}+2^{2008}+...+2+2^0\)
\(=1+2+...+2^{2008}+2^{2009}\)
\(\Rightarrow2A=2+2^2+...+2^{2010}\)
\(\Rightarrow2A-A=\left(2+2^2+...+2^{2010}\right)-\left(1+2+...+2^{2009}\right)\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)\)
\(=2^{2010}-2^{2010}+1=1\)
Vậy M = 1
1.
M = 22010 - ( 22009 + 22008 + ... + 21 + 20 )
đặt N = 22009 + 22008 + ... + 21 + 20
2N = 22010 + 22009 + ... + 22 + 21
2N - N = ( 22010 + 22009 + ... + 22 + 21 ) - ( 22009 + 22008 + ... + 21 + 20 )
N = 22010 - 20
Thay N vào ta được :
M = 22010 - ( 22010 - 20 )
M = 22010 - 22010 + 20
M = 20 = 1
2.
Ta có :
2332 < 2333 = ( 23 ) 111 = 8111
3223 > 3222 = ( 32 ) 111 = 9111
Vì 2332 < 8111 < 9111 < 3223