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a) \(\frac{5}{2.m}=\frac{1}{6}+\frac{n}{3}\) \(\left(m\ne0\right)\)
\(\frac{15}{6.m}=\frac{m}{6.m}+\frac{2.m.n}{6.m}\)
\(\frac{15}{6.m}=\frac{m+2mn}{6.m}\)
\(m+2mn=15\)
\(m\left(1+2n\right)=15\)
\(\Rightarrow m\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)
Với m = 1, 1 + 2n = 15 hay n = 7.
Với m = 3, 1 + 2n = 5 hay n = 2
Với m = 5, 1 + 2n = 2 hay n = 1
Với m = 15, 1 + 2n = 1 hay n = 0.
Vậy ta tìm được 4 cặp (m;n) thỏa mãn là: (1;7) , (3;2) , (5;1) và (15;0)
Câu b, c hoàn toàn tương tự.
\(\frac{m}{2}-\frac{2}{n}=\frac{1}{2}\)
\(\Rightarrow\frac{2}{n}=\frac{m}{2}-\frac{1}{2}\)
\(\Rightarrow\frac{2}{n}=\frac{m-1}{2}\)
\(\Rightarrow\hept{\begin{cases}2=m-1\\n=2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}m=3\\n=2\end{cases}}\)
Câu còn lại làm nốt
\(\frac{m}{2}-\frac{2}{n}=\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{n}=\frac{m}{2}-\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{n}=\frac{m-1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}2=m-1\\n=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}m=3\\n=2\end{cases}}\)
\(\frac{1}{m}-\frac{n}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{n}{6}=\frac{1}{m}-\frac{1}{2}\)
\(\Leftrightarrow\frac{n}{6}=\frac{2-m}{2m}\)
\(\Leftrightarrow\orbr{\begin{cases}n=2-m\\6=2m\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}n=2-m\\m=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}n=2-3\\m=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}n=-1\\m=3\end{cases}}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\) <=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+.....\frac{1}{n}-\frac{1}{n-1}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)
<=>\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
<=>\(\frac{1}{n+1}=\frac{1}{2001}\)
<=>n+1 =2001
<=>n = 2000
ta có:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2.3}+\frac{1}{2.6}+\frac{1}{2.10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2001}\)
=>\(n+1=2001\)
=>\(n=2000\)
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 bang 3
n bằng 2
Ta có \(\frac{1}{m}\)+\(\frac{n}{6}\)=\(\frac{1}{2}\)
\(\frac{1}{m}\)=\(\frac{1}{2}\)-\(\frac{n}{6}\)
\(\frac{1}{m}\)=\(\frac{3}{6}\)-\(\frac{n}{6}\)
\(\frac{1}{m}\)=\(\frac{3-n}{6}\)
=>m*(3-n)=6
=>3-nEƯ(6)
Ta có bảng giá trị