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a) 20+ 21+22+...+22010
A= 20+ 21+22+...+22010
2A= 2( 20+ 21+22+...+22010)
2A= 21+22+...+22010+22011
2A-A= (21+22+...+22010+22011) -(20+ 21+22+...+22010)
A= 22011-20
A= 22011-1
Vì 22011 > 22010 nên 22011 -1 > 22010-1
Vậy..
c)1030 = ( 103 )10 = 100010
= ( 210 )10 = 102410
Vì 1024 > 1000
=> 100010 < 102410 hay 1030 < 2100
Ta có : \(A=2^0+2^1+2^2+2^3+...+2^{2010}\)
\(3A=2+2^2+2^3+2^4+...+2^{2011}\)
=> \(2A=3A-A=\left(2^1+2^2+...+2^{2011}\right)-\left(2^0+2^1+...+2^{2010}\right)\)
=>\(2A=2^{2011}-1\)
=>\(A=\frac{2^{2011}-1}{2}\)
=> A < B ( vì \(\frac{2^{2011}-1}{2}< 2^{2011}\) )
Bài 1:
a. https://olm.vn/hoi-dap/detail/100987610050.html
b. Giống nhau hoàn toàn => P=Q
Chỉ biết thế thôi
\(A=2^0+2^1+2^2+...+2^{2010}=1+2+2^2+...+2^{2010}\)
\(A=1+2\left(2^0+2^1+2^2+...+2^{2019}\right)=1+2\left(A-2^{2010}\right)=1+2A-2^{2011}\)
\(A=2^{2011}-1=B\)
a) Ta có :
\(A=\frac{10^{2010}+1}{10^{2011}+1}\)
\(\Rightarrow10A=\frac{10^{2011}+10}{10^{2011}+1}=\frac{\left(10^{2011}+1\right)+9}{10^{2011}+1}=1+\frac{9}{10^{2011}+1}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}\)
\(\Rightarrow10B=\frac{10^{2012}+10}{10^{2012}+1}=\frac{\left(10^{2012}+1\right)+9}{10^{2012}+1}=1+\frac{9}{10^{2012}+1}\)
Vì \(\frac{9}{10^{2011}+1}>\frac{9}{10^{2012}+1}\)nên \(10A>10B\)
\(\Rightarrow A>B\)
Vậy : \(A>B\)
b) Ta có :
\(\left(\frac{-1}{2}\right)^{11}=\frac{-1^{11}}{2^{11}}=\frac{-1}{2^{11}}\)
\(\left(\frac{-1}{2}\right)^{13}=\frac{-1^{13}}{2^{13}}=\frac{-1}{2^{13}}\)
Vì \(\frac{-1}{2^{11}}>\frac{-1}{2^{13}}\)nên \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)
Vậy : \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+1+9}{10^{2012}+1+9}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+10}{10^{2012}+10}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10\cdot\left(10^{2010}+1\right)}{10\cdot\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)
Vậy : B < A
a, 2A = 2+2^2+....+2^2012
A=2A-A=(2+2^2+....+2^2012)-(1+2+2^2+....+2^2011) = 2^2012-1 > 2^2011-1 = B
=> A>B
b, A = 2009.(2010+1) = 2009.2010+2009 = (2009.2010+2010)-1 = 2010.(2009+1)-1 = 2010.2010-1 = 2010^2-1 < 2010^2 = B
=> A<B
c, A = (10^3)^10 = 1000^10 < 1024^10 = (2^10)^10 = 2^100 = B
=> A<B
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