\(A=\frac{1010^{1010}}{2010^{2010}},B=\frac{2010^{2010}}{3010^{3010}}\)
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Vì 1000/2009>1000/2009+2010 (1)
1010/2010>1010/2009+2010 (2)
Ta cộng theo vế (1) và (2) với nhau nên ta được:
1000/2009+1010/2010>1000/2009+2010 +1010/2009+2010
=>1000/2009+1010/2010>1000+1010/2009+2010
Vậy A<B
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Ta có: \(3^{2010}=3^{10}\cdot3^{2000}=3^{10}\cdot9^{1000}\)
\(2^{3010}=2^{10}\cdot2^{3000}=2^{10}\cdot8^{1000}\)
Xét thấy \(3^{10}>2^{10};9^{1000}>8^{1000}\)
\(\Rightarrow3^{10}\cdot9^{1000}>2^{10}\cdot8^{1000}\)
Vậy \(3^{2010}>2^{3010}\)
Ta có B=\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<1
=>\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<\(\frac{2009^{2010}-2+3}{2009^{2011}-2+3}\)=\(\frac{2009^{2010}+1}{2009^{2011}+1}\)(1)
Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<1
=> \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<\(\frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}\)=\(\frac{2009^{2010}+2009}{2009^{2011}+2009}\)=\(\frac{2009\cdot\left(2009^{2009}+1\right)}{2009\cdot\left(2009^{2010}+1\right)}\)=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)=A(2)
Từ (1)và(2)=>B<\(\frac{2009^{2010}+1}{2009^{2011}+1}\)<A=>B<A hay A>B
\(\frac{2010\cdot2011+1000}{2012\cdot2010-1010}\)
= \(\frac{2010\cdot2011+1000}{\left(2011+1\right)\cdot2010-1010}\)
= \(\frac{2010\cdot2011+1000}{2011\cdot2010+2010-1010}\)
= \(\frac{2010\cdot2011+1000}{2011\cdot2010+1000}\)
= 1
\(\frac{2010.2011+1000}{2012.2010-1010}\)
\(=\frac{2010.2011+2010-1010}{2012.2010-1010}\)
\(=\frac{2010.\left(2011+1\right)-1010}{2012.2010-1010}\)
\(=\frac{2010.2012-1010}{2012.2010-1010}\)
\(=1\)
\(\frac{2010.125+1000}{126.2010-1010}=\frac{10\left(201.125+100\right)}{10\left(201.126-101\right)}=\frac{201.125-101+201}{2011.126-101}\)
\(=\frac{201.126-101}{201.126-101}=1\)
=2010 x 125 + 2010x 126 + 1000+1010
=2010 x 125 + 2010 x 126 + 2010
=2010 x ( 125 + 126 +1 )
=2010 x 252
=506520
Bổ sung đề: So sánh A và B
Ta có:
A. \(2010^{1000}=\frac{1010^{1010}.2010^{1000}}{2010^{2010}}=\left(\frac{101}{201}\right)^{1010}\)
B. \(2010^{1000}=\frac{2010^{2010}.2010^{1000}}{3010^{3010}}=\left(\frac{201}{301}\right)^{3010}\)
Từ \(\frac{101}{201}>\frac{1}{2}>\frac{40401}{90601}=\left(\frac{201}{301}\right)^2\)và \(\frac{201}{301}< 1\)
có: \(\left(\frac{101}{201}\right)^{1010}>\left(\frac{201}{301}\right)^{2.1010}=\left(\frac{201}{301}\right)^{2020}>\left(\frac{201}{301}\right)^{3010}\)
Suy ra \(A=\left(\frac{101}{201}\right)^{1010}.\frac{1}{2010^{1000}}>\left(\frac{201}{301}\right)^{3010}.\frac{1}{2010^{1000}}\) hay A > B