(2\(\frac{3}{5}\cdot\frac{1111}{2011}-\frac{2222}{2011}\)) :\(\frac{2222}{8044}\)=?
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17 tháng 10 2018
\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)
\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)
Vậy \(\frac{B}{A}=2012\)
Chúc bạn học tốt ~
16 tháng 9 2017
\(\frac{1010+1111+1212+1313+1414+1515+1616+1717}{2020+2121+2222+2323+2424+2525+2626+2727}\)
\(=\frac{101.10+101.11+...+101.17}{101.20+101.21+...+101.27}\)
\(=\frac{101.\left(10+11+...+17\right)}{101.\left(20+21+...+27\right)}\)
\(=\frac{108}{188}\)
\(=\frac{27}{47}\)
16 tháng 9 2017
\(2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right)\cdot5.y>\frac{5}{6}\)
\(\Rightarrow2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right):5.y>\frac{5}{6}\)
\(\Rightarrow2>\left(\frac{20}{120}+\frac{16}{120}+\frac{9}{120}+\frac{5}{120}\right):5.y>\frac{5}{6}\)
\(\Rightarrow2>\frac{5}{12}:5.y>\frac{5}{6}\)
\(\Rightarrow2>\frac{1}{12}.y>\frac{5}{6}\)
Đặt :\(\frac{1}{12}.y=2\Rightarrow y=2:\frac{1}{12}=24\)
\(\frac{1}{12}.y=\frac{5}{6}\Rightarrow y=\frac{5}{6}:\frac{1}{12}=10\)
\(\Rightarrow24>y>10\)
\(\Rightarrow y\in\left\{11;12;...;23\right\}\)
Y
13 tháng 6 2019
2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
Y
13 tháng 6 2019
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
\(\left(\frac{13}{5}x\frac{11}{20}-\frac{11}{10}\right):\frac{11}{40}=\left(\frac{143}{100}-\frac{110}{100}\right):\frac{11}{40}=\frac{33}{100}:\frac{33}{120}=\frac{33}{100}x\frac{120}{33}=\frac{6}{5}\)
thank gì cho đúng đi