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D=\(\frac{2011^{2013}+1}{2011^{2014}+1}\)
<\(\frac{2011^{2013}+1+2010}{2011^{2014}+1+2010}\)
<\(\frac{2011^{2013}+2011}{2011^{2014}+2011}\)
<\(\frac{2011\left(2011^{2012}+1\right)}{2011\left(2011^{2013}+1\right)}\)
<\(\frac{2011^{2012}+1}{2011^{2013}+1}\)
<C
Vậy C>D
TA CÓ :
\(B=\frac{2010+2011+2012}{2011+2012+2013}\)
\(B=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
VÌ : \(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
=> A > B
VẬY , A > B
Mình tự hỏi. sao banh biết rồi còn đăng lên làm gì??????????
a,ta có:
A=2011^2012-2011^2011
=2011^2011(2011-1)
=2011^2011.2010
và B = 2011^2013-2011^2012
=2011^2012(2011-1)
=2011^2012.2010
Vì 2011^2011<2011^2012 => 2011^2011.2010< 2011^2012.2010
=>A<B
a,ta có:
A=2011^2012-2011^2011
=2011^2011(2011-1)
=2011^2011.2010
và B = 2011^2013-2011^2012
=2011^2012(2011-1)
=2011^2012.2010
Vì 2011^2011<2011^2012 => 2011^2011.2010< 2011^2012.2010
=>A<B
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)
\(<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(<1-\frac{1}{2010}\)
\(<\frac{2009}{2010}<1\)
=>N<1
P=\(\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2014-1}{2014}+\frac{2015-1}{2015}\)
=\(1-\frac{1}{2012}+1-\frac{1}{2013}+1-\frac{1}{2014}+1-\frac{1}{2015}\)
=\(4-\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)\)
VẬY P<4
\(\orbr{\begin{cases}\orbr{\begin{cases}\frac{2011}{2012}< 1\\\frac{2012}{2013}< 1\end{cases}}\\\orbr{\begin{cases}\frac{2013}{2014}< 1\\\frac{2014}{2015}< 1\end{cases}}\end{cases}\Rightarrow P< 1+1+1+1=4}\)
\(P=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}\)
\(\Rightarrow P>\frac{2012}{2013}+\frac{2012}{2013}+\frac{2012}{2013}\)
\(P>\frac{4036}{2013}>1\)(1)
\(Q=\frac{2010+2011+2012}{2011+2012+2013}=\frac{6033}{6036}< 1\)(2)
\(Q< 1;P>1\Rightarrow P>Q\)
Câu hỏi của Son Goku - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo bài bạn Huy nhé!
a) \(\frac{2^{10}+1}{2^{10}-1}\)và \(\frac{2^{10}-1}{2^{10}-3}\)
Ta có chính chất phân số trung gian là \(\frac{2^{10}+1}{2^{10}-3}\)
\(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}\) ; \(\frac{2^{10}-1}{2^{10}-3}< \frac{2^{10}+1}{2^{10}-3}\)
Vì \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}>\frac{2^{10}-1}{2^{10}-3}\)
Nên \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}-1}{2^{10}-3}\)
b) \(A=\frac{2011}{2012}+\frac{2012}{2013}\)và \(B=\frac{2011+2012}{2012+2013}\)
Ta có : \(A=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}=B\)
Vậy A > B
Có gì sai cho sorry
a,
\(\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}=\frac{2^{10}-1}{2^{10}-3}\)
b,
\(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
Sửa lại:
Ta có:
\(2011A=\frac{2011^{2013}+2011}{2011^{2013}+1}=1+\frac{2010}{2011^{2013}+1}\)
\(2011B=\frac{2011^{2014}+2011}{2011^{2014}+1}=1+\frac{2010}{2011^{2014}+1}\)
Vì \(1+\frac{2010}{2011^{2013}+1}>1+\frac{2010}{2011^{2014}+1}\) nên 2011A > 2011 B
Từ đó A > B
Vậy A > B
Có:
\(2009A=\frac{2011^{2013}+2011}{2011^{2013}+1}=1+\frac{2010}{2011^{2013}+1}\)
\(2011B=\frac{2011^{2014}+2011}{2011^{2014}+1}=1+\frac{2010}{2011^{2014}+1}\)
Mà \(1+\frac{2010}{2011^{2013}+1}>1+\frac{2010}{2011^{2014}+1}\)
\(\Rightarrow2009A>2009B\)
\(\Rightarrow A>B\)
Vậy A > B