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\(\frac{m}{2}-\frac{2}{n}=\frac{1}{2}\)
\(\Rightarrow\frac{2}{n}=\frac{m}{2}-\frac{1}{2}\)
\(\Rightarrow\frac{2}{n}=\frac{m-1}{2}\)
\(\Rightarrow\hept{\begin{cases}2=m-1\\n=2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}m=3\\n=2\end{cases}}\)
Câu còn lại làm nốt
\(\frac{m}{2}-\frac{2}{n}=\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{n}=\frac{m}{2}-\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{n}=\frac{m-1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}2=m-1\\n=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}m=3\\n=2\end{cases}}\)
\(\frac{1}{m}-\frac{n}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{n}{6}=\frac{1}{m}-\frac{1}{2}\)
\(\Leftrightarrow\frac{n}{6}=\frac{2-m}{2m}\)
\(\Leftrightarrow\orbr{\begin{cases}n=2-m\\6=2m\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}n=2-m\\m=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}n=2-3\\m=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}n=-1\\m=3\end{cases}}\)
Ta có: \(M=\frac{2014^2+1^2}{2014.1}+\frac{2013^2+2^2}{2013.2}+\frac{2012^2+3^2}{2012.3}+...+\frac{1008^2+1007^2}{1008.1007}\)
\(=\frac{2014}{1}+\frac{1}{2014}+\frac{2013}{2}+\frac{2}{2013}+\frac{2012}{3}+\frac{3}{2013}+...+\frac{1008}{1007}+\frac{1007}{1008}\)
\(=\frac{2014}{1}+\frac{2013}{2}+...+\frac{1}{2014}\)
\(=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)\)
\(=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+\frac{2015}{2015}\)
\(=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}}=2015\)
Sửa N=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)
Ta có : \(\frac{1}{2}< \frac{2}{3}\); \(\frac{3}{4}< \frac{4}{5}\); \(\frac{5}{6}< \frac{6}{7}\); ... ; \(\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)hay M < N
b) M .N = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}=\frac{1.2.3.4.5.6...99.100}{2.3.4.5.6.7...100.101}=\frac{1}{101}\)
c) vì M < N nên M. M < M . N = \(\frac{1}{101}\)\(< \frac{1}{100}\)
\(\Rightarrow M< \frac{1}{10}\)
\(M=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+....+\frac{1}{2^{43}}-\frac{1}{2^{46}}+\frac{1}{2^{49}}-\frac{1}{2^{52}}\)
Nên \(2^3.M=4-\frac{1}{2}+\frac{1}{2^4}-\frac{1}{2^7}+.....+\frac{1}{2^{46}}-\frac{1}{2^{52}}\)
Suy ra \(2^3.M-M=4-\frac{1}{2^{52}}\)hay\(7.M=4-\frac{1}{2^{52}}\).
Khi đó \(M=\frac{4}{7}-\frac{1}{2^{52}.7}< 1\)
Vì \(\frac{9}{4}>1;M< 1\)nên \(\frac{9}{4}>M\)
Vậy \(\frac{9}{4}>M\)