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Ta có :
\(\frac{1}{2013}M=\frac{2013^{2012}+2012}{2013^{2012}+2013}=\frac{2013^{2012}+2013}{2013^{2012}+2013}-\frac{1}{2013^{2012}+2013}=1-\frac{1}{2013^{2012}+2013}\)
Lại có :
\(\frac{1}{2013}N=\frac{2013^{2011}+2012}{2013^{2011}+2013}=\frac{2013^{2011}+2013}{2013^{2011}+2013}-\frac{1}{2013^{2011}+2013}=1-\frac{1}{2013^{2011}+2013}\)
Vì \(\frac{1}{2013^{2012}+2013}< \frac{1}{2013^{2011}+2013}\) nên \(M=1-\frac{1}{2013^{2012}}>N=1-\frac{1}{2013^{2011}+2013}\)
Vậy \(M>N\)
Chúc bạn học tốt ~
Biển Cửa Lò, chùa Thiên mụ, núi Ngũ Hành Sơn, chùa Cầu Hội An, kinh thành Huế, đèo Hải Vân
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Ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{1006}\)
\(=\frac{1}{1007}+\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2012}+\frac{1}{2013}\left(1\right)\)
Mà \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow S=P\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
Vậy...
Mọi người tk mình đi mình đang bị âm nè!!!!!!
Ai tk mình mình tk lại nha !!!
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}.....+\frac{1}{2012}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-........-\frac{1}{1006}\)
\(S=\frac{1}{1007}+\frac{1}{1008}+.......+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}\)
\(\Rightarrow\left(S-P\right)^2=\left(\frac{1}{1007}+\frac{1}{1008}+....+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{1007}-\frac{1}{1008}-....-\frac{1}{2012}-\frac{1}{2013}\right)^2\)
\(\Rightarrow\left(S-P\right)^2=0\)
Vậy \(\left(S-P\right)^2=0\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)
\(S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)
\(\Rightarrow S-P=0\)
\(\Rightarrow\left(S-P\right)^{2013}=0\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(=\left(1+\frac{1}{3}+......+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+........+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+......+\frac{1}{2013}\)
\(=P\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0\)
1/ (69.210+1210)+(219.273+15.49.94) = 29.39.210+310.220+219.39+5.3.218.38 = 219.39+310.220+219.39+5.218.39
= 218.39(2+3.22+5)=19.218.39
\(\frac{B}{A}=\frac{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}\)
\(=\frac{\left(\frac{2011}{2}+1\right)+\left(\frac{2010}{3}+1\right)+...+\left(\frac{1}{2012}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}\)
\(=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+....+\frac{2013}{2012}+\frac{2013}{2013}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}=2013\)
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