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a = 2011.2013
a = 2011.(2012+1)
a = 2011.2012 + 2011
b = 2012.2012
b = (2011+1).2012
b = 2011.2012 + 2012
Vì 2011 < 2012
=> 2011.2012 + 2011 < 2011.2012 + 2012
=> a < b
\(10A=\dfrac{10^{2021}+10}{10^{2021}+1}=\dfrac{\left(10^{2021}+1\right)+9}{10^{2021}+1}=\dfrac{10^{2021}+1}{10^{2021}+1}+\dfrac{9}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=\dfrac{\left(10^{2022}+1\right)+9}{10^{2022}+1}=\dfrac{10^{2022}+1}{10^{2022}+1}+\dfrac{9}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
Vì \(10^{2022}>10^{2021}=>10^{2021}+1< 10^{2022}+1\)
\(=>\dfrac{9}{10^{2021}+1}>\dfrac{9}{10^{2022}+1}\)
\(=>10A>10B\)
\(=>A>B\)
Ta có: A = 32 x 53 - 31 = 31 x 53 + 53 - 31 = 31 x 53 + 22
Do 22 < 32 => 31 x 53 + 22 < 53 x 31 + 32
=> 52 x 53 - 31 < 53 x 31 + 32
=> A < B
\(A=32\times53-31\)
\(\Leftrightarrow A=\left(31+1\right)\times53-31\)
\(\Leftrightarrow A=31\times53+53-31\)
\(\Leftrightarrow A=31\times53+22\)
Vì 22 < 32 nên 31 x 53 + 22 < 53 x 31 + 32 hay A < B
~ Hok tốt ~
Ta có: \(B=2020.2021.2022=\left(2021-1\right).\left(2021+1\right).2021=\left(2021-1\right)^2.2021< 2021^2.2021=A\)
a= 202 x 204 = 202 x (203+1)=202 x 203 + 202
b=203 x 203 = (202+1) x 203 = 202 x 203 + 203
Vì 203>202 => 202x 203 + 202<202x203 +203
=>a<b
A = 2006.2007+2008
A = (2008 - 2).2007 + 2008
A = 2008.2007 - 2.2007 + 2008
A = 2008.2007 - 4014 + 2008
A = 2008.2007 - 2006 = B
Vậy A = B
Gọi số 1987656 là a
Ta có:
\(A=\left(a+1\right).\left(a-1\right)\)
\(A=a^2-1+a-1\)
\(A=a^2+a-\left(1-1\right)\)
\(A=a^2+a\)
\(B=a.a\)
\(B=a^2\)
Vì \(a^2+a>a^2\Rightarrow A>B\)
A=(1987656+1). 1987655
A=1987656.1987655+1987655
B=(1987655+1).1987656+1987656
suy ra 1978656.1987655=1987656 và 1987655<1987656 nên A<B