Do  a < b < c < d < m < n 
=> 2c < c + d 
m< n => 2m < m+ n 
=> 2c + 2a +2m = 2 ( a + c + m) < a +b + c + d + m + n) 
Do đó :
\(\dfrac{\text{(a + c + m)}}{\left(a+b+c+d+m+n\right)}\) < \(\dfrac{1}{2}\)

Do a < b < c < d < m < n

=> a + c + m < b + d + n

=> 2 × (a + c + m) < a + b + c + d + m + n

=> a + c + m / a + b + c + d + m + n < 1/2 ( đpcm)

Do a < b < c < d < m < n

=> a + c + m < b + d + n

=> 2 × (a + c + m) < a + b + c + d + m + n

=> a + c + m / a + b + c + d + m + n < 1/2 ( đpcm)

Do a<b<c<d<m<n

\(\Rightarrow\)a+c+m<b+d+n

\(\Rightarrow\)2(a+c+m) < a+b+c+d+m+n

\(\Rightarrow\)\(\frac{2\left(a+c+m\right)}{a+b+c+d+m+n}< 1\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)

k cho mình nha

Do a < b < c < d < m < n

=> a + c + m < b + d + n

=> 2.(a + c + m) < a + b + c + d + m + n

=> \(\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\) (đpcm)

ai làm đk mình k cho

Ta có:  a < b     =>    2a < a + b

           c < d      =>    2c < c + d

           m < n     =>    2m < m +n

suy ra:    2 ( a + c + m)  < a + b + c + d + m + n

=>   \(\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)

Vì a < b

     c < d

     m < n

=> a + c + m < b + d + n

=> 2 ( a + c + m ) < b + d + n + a + c + m

=> \(\frac{a+c+m}{2\left(a+c+m\right)}\)\(>\)\(\frac{a+c+m}{a+b+c+d+m+n}\)

=> \(\frac{1}{2}>\frac{a+c+m}{a+b+c+d+m+n}\)

\(\hept{\begin{cases}a< b\Rightarrow2a< a+b\\c< d\Rightarrow2c< c+d\\m< n\Rightarrow2m< m+n\end{cases}}\)

\(\Rightarrow2\left(a+c+m\right)< a+b+c+d+m+n\)

\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\left(đpcm\right)\)

Do  a < b < c < d < m < n 
=> 2c < c + d 
m< n => 2m < m+ n 
=> 2c + 2a +2m = 2 ( a + c + m) < a +b + c + d + m + n) 
Do đó :
(a + c + m)/(a + b + c + d + m + n) < 1/2(đcpcm)

Vì \(a< b< c< d< m< n\)

\(\Rightarrow\hept{\begin{cases}a+c+m< 3a\\a+b+c+d+m+n< 6a\end{cases}}\)

\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{3a}{6a}\)

\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\left(đpcm\right)\)

                                                             Bài giải

Ta có : \(a< b\text{ }\Rightarrow\text{ }2a< a+b\)

        \(c< d\text{ }\Rightarrow\text{ }2c< c+d\)

         \(m< n\text{ }\Rightarrow\text{ }2m< m+n\)

\(\Rightarrow\text{ }2a+2c+2m< \left(a+b+c+d+m+n\right)\) \(\Leftrightarrow\text{ }2\left(a+c+m\right)< \left(a+b+c+d+m+n\right)\)

\(\Rightarrow\text{ }\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)

Trong vở Bài tập nâng cao và một số chuyên đề Toán 7 trang đầu đó bạn !

                                                             Bài giải

Ta có : \(a< b\text{ }\Rightarrow\text{ }2a< a+b\)

        \(c< d\text{ }\Rightarrow\text{ }2c< c+d\)

         \(m< n\text{ }\Rightarrow\text{ }2m< m+n\)

\(\Rightarrow\text{ }2a+2c+2m< \left(a+b+c+d+m+n\right)\) \(\Leftrightarrow\text{ }2\left(a+c+m\right)< \left(a+b+c+d+m+n\right)\)

\(\Rightarrow\text{ }\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)

ta có 

a<b<c=>3a<a+b+c

d<m<n=>3d<d+m+n

=>3a+3d<a+b+c+d+m+n

=>3a+3a/a+b+c+d+m+n<a+b+c+m+n+d/a+b+c+d+m+n

=>3(a+d)/a+b+c+d+m+n)<1

=>a+d/a+b+c+d+m+n<1/3  (đpcm)