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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 ~
A = 1×2 + 2×3 + 3×4 + ... + 98×99
3A = 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 98×99×(100-97)
3A = 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 98×99×100 - 97×98×99
3A = 98×99×100
A = 98×33×100
A = 323400
2) Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
Ta có:
102012 + 1/102013 + 1 < 102012 + 1 + 9/102013 + 1 + 9
< 102012 + 10/102013 + 10
< 10.(102011 + 1)/10.(102012 + 1)
< 102011 + 1/102012 + 1
Vào lúc: 2016-07-17 13:22:30 Xem câu hỏi
1) A = 1×2 + 2×3 + 3×4 + ... + 98×99
3A = 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 98×99×(100-97)
3A = 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 98×99×100 - 97×98×99
3A = 98×99×100
A = 98×33×100
A = 323400
2) Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
Ta có:
102012 + 1/102013 + 1 < 102012 + 1 + 9/102013 + 1 + 9
< 102012 + 10/102013 + 10
< 10.(102011 + 1)/10.(102012 + 1)
< 102011 + 1/102012 + 1
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
So sánh:
\(A=-\frac{9}{10^{2012}}-\frac{19}{10^{2011}}\) và \(B=-\frac{9}{10^{2011}}-\frac{19}{10^{2012}}\)
Ta có:
\(A=-\frac{9}{10^{2012}}-\frac{19}{10^{2011}}=-\frac{1}{10^{2011}}\left(\frac{9}{10}+19\right)=-\frac{1}{10^{2011}}.\frac{199}{10}\)
\(B=-\frac{9}{10^{2011}}-\frac{19}{10^{2012}}=-\frac{1}{10^{2011}}\left(9+\frac{19}{10}\right)=-\frac{1}{10^{2011}}.\frac{109}{10}\)
Vì \(\frac{199}{10}>\frac{109}{10}\Rightarrow\frac{1}{10^{2011}}.\frac{199}{10}>\frac{1}{10^{2011}}.\frac{109}{10}\Rightarrow-\frac{1}{10^{2011}}.\frac{199}{10}< -\frac{1}{10^{2011}}.\frac{109}{10}\)
Vậy nên A<B
\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)......\left(1-\frac{1}{2011}\right)\)
\(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2010}{2011}\)
\(=\frac{6.7.8.9.....2010}{7.8.9.10.....2011}\)
\(=\frac{6}{2011}\)
N =\(\frac{2010+2011+2012}{2011+2012+2013}\)
\(\Rightarrow N=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
Do: \(\frac{2010}{2011}>\frac{2010}{2011+2012+2013};\frac{2011}{2012}>\frac{2011}{2011+2012+2013};\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
\(\Rightarrow\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
\(\Rightarrow\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010+2011+2012}{2011+2012+2013}\Leftrightarrow N>M\)
\(B< \frac{10^{2012}+1+9}{10^{2013}+1+9}=\frac{10^{2012}+10}{10^{2013}+10}=\frac{10\left(10^{2011}+1\right)}{10\left(10^{2012}+1\right)}=\frac{10^{2011}+1}{10^{2012}+1}=A\)
Vậy A > B
Áp dụng bất đẳng thức :
\(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\)
Ta có :
\(B=\frac{10^{2012}+1}{10^{2013}+1}< \frac{10^{2012}+1+9}{10^{2013}+1+9}=\frac{10^{2012}+10}{10^{2013}+10}=\frac{10\left(10^{2011}+1\right)}{10\left(10^{2012}+1\right)}=\frac{10^{2011}+1}{10^{2012}+1}=A\)
\(\Leftrightarrow B< A\)