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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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\(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)
Bài giải
Theo bài ra :
\(A=\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}\)
\(B=\frac{2009+2010+2011}{2010+2011+2012}=\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)
Ta có :
\(\frac{2009}{2010}>\frac{2009}{2010+2011+2012}\)
\(\frac{2010}{2011}>\frac{2010}{2010+2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2010+2011+2012}\)
\(\Rightarrow\text{ }\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}>\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)
\(\Rightarrow\text{ }A>B\)
\(1-A=1-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010^{2012}+1}{2010^{2012}+1}-\frac{2010^{2011}+1}{2010^{2012}+1}\)=\(\frac{2010}{2010^{2012}+1}\)
\(1-A=1-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010^{2012}+1}{2010^{2012}+1}-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010}{2010^{2012}+1}\)
\(1-B=1-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010^{2011}+1}{2010^{2011}+1}-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010}{2010^{2011}+1}\)
\(\frac{2010}{2010^{2012}+1}<\frac{2010}{2010^{2011}+1}\Rightarrow A>B\)
so sánh : cho A\(\frac{2010^{2011}+1}{2010^{2012}+1}\)
cho B =\(\frac{2010^{2010}+1}{2010^{2011}+1}\)
Ta có:
\(A=\frac{2010^{2011}+1}{2010^{2012}+1}\)
\(2010A=\frac{2010^{2012}+2010}{2010^{2012}+1}\)
\(2010A=1+\frac{2009}{2010^{2012}+1}\)
Lại có:
\(B=\frac{2010^{2010}+1}{2010^{2011}+1}\)
\(2010B=\frac{2010^{2011}+2010}{2010^{2011}+1}\)
\(2010B=1+\frac{2009}{2010^{2011}+1}\)
Vì \(1+\frac{2009}{2010^{2012}+1}< 1+\frac{2009}{2010^{2011}+1}\)
nên 2010A < 2010B
hay A < B
Vậy A < B
\(1-A=1-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010^{2012}+1}{2010^{2012}+1}-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010}{2010^{2012}+1}\)
\(1-B=1-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010^{2011}+1}{2010^{2011}+1}-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010}{2010^{2011}+1}\)
Do \(\frac{2010}{2010^{2012}+1}<\frac{2010}{2010^{2011}+1}\)nên \(A>B\)
Do 20102011+1<20102012+1=>A<1
Tương tự với B;B<1
Theo đề bài ta có:
\(A=\frac{2010^{2011}+1}{2010^{2012}+1}<\frac{2010^{2011}+1+2009}{2010^{2012}+1+2009}=\frac{2010^{2011}+2010}{2010^{2012}+2010}=\frac{2010.\left(1+2010^{2010}\right)}{2010.\left(1+2010^{2011}\right)}=\frac{2010^{2010}+1}{2010^{2011}+1}=B\)(*)
Từ (*)=> A<B
A=\(\frac{-199}{10^{2011}}\)
B=\(\frac{-109}{10^{2011}}\)
Dễ dàng so sánh được A<B
+ta có 10^2010=10...0(2010 số 0)
và 10^2011=10...0(2011 số 0)
suy ra -9/10...0(2010 số 0)= -90/10...0(2011 số 0)[nhân tử,mẫu cho 10]
suy ra A=-90/10...0(2011 số 0)+-19/10...0(2011 số 0)= -109/10...0(2011 số 0) [1]
+-19/10...0(2010 số 0)= -190/10...0(2011 số 0)[nhân tử,mẫu cho 10]
và 10^2011=10...0(2011 số 0)
suy ra -9/10...0(2011 số 0)+-190/10...0(2011 số 0)= -199/10...0(2011 số 0) [2]
vì -109>-199 suy ra [1]>[2]
K CHO MIK VS BẠN ƠIIIIIIIIIIIIIIIIIII
\(-A=\frac{9}{10^{2010}}+\frac{19}{10^{2011}}\)
\(-A=\frac{9}{10^{2010}}+\frac{10}{10^{2011}}+\frac{9}{10^{2011}}\)
\(-A=\frac{9}{10^{2010}}+\frac{1}{10^{2010}}+\frac{9}{10^{2011}}\)
\(-A=\frac{10}{10^{2010}}+\frac{9}{10^{2011}}\)
\(-A=\frac{1}{10^{2009}}+\frac{9}{10^{2011}}\)
\(-B=\frac{9}{10^{2011}}+\frac{19}{10^{2010}}\)
Làm tương tự nhé
ta thấy -b > -a nên a>b