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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 : \(a< b\text{ }\Rightarrow\text{ }2a< a+b\)
\(c< d\text{ }\Rightarrow\text{ }2c< c+d\)
\(m< n\text{ }\Rightarrow\text{ }2m< m+n\)
\(\Rightarrow\text{ }2a+2c+2m< \left(a+b+c+d+m+n\right)\) \(\Leftrightarrow\text{ }2\left(a+c+m\right)< \left(a+b+c+d+m+n\right)\)
\(\Rightarrow\text{ }\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)
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)
Từ:\(\hept{\begin{cases}a< c\\c< d\\m< n\end{cases}}\Rightarrow a+c+m< c+d+n\)
\(\Rightarrow2\left(a+c+n\right)< a+b+c+d+m+n\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)
\(\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{a+c+m}{a+c+m+a+c+m}\)
\(=\dfrac{a+c+m}{2a+2c+2m}=\dfrac{1}{2}\) ( do a < b < c < d < m < n )
\(\Rightarrowđpcm\)
\(\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{a+c+m}{a+b+c+a+b+c}\left(a< b< c< d< m< n\right)\)\(\Rightarrow\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{a+c+m}{2a+2c+2m}\)
\(\Rightarrow\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{a+c+m}{2\left(a+c+m\right)}\)
\(\Rightarrow\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{1}{2}\rightarrowđpcm\)
Ta có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< bc\)
\(\Leftrightarrow2018ad< 2018bc\)
\(\Leftrightarrow2018ad+cd< 2018bc+cd\)
\(\Leftrightarrow d\left(2018a+c\right)< c\left(2018b+d\right)\)
\(\Leftrightarrow\frac{2018a+c}{2018b+d}< \frac{c}{d}\left(đpcm\right)\)
Ta có: a/(a+b) > a/(a+b+c)
b/(b+c) > b/(b+c+a)
c/(c+a) > c/(c+a+b)
=> [a/(a+b)] + [b/(b+c)] + [c/(c+a)] > [a/(a+b+c)] + [b/(a+b+c)] + [c/(a+b+c)]
=> [a/(a+b)] + [b/(b+c)] + [c/(c+a)] > 1
Lại có: a/(a+b) < (a+b)/(a+b+c)
b/(b+c) < (b+c)/(b+c+a)
c/(c+a) < (c+a)/(c+a+b)
=> [a/(a+b)] + [b/(b+c)] + [c/(c+a)] < [(a+b)/(a+b+c)] + [(b+c)/(a+b+c)] + [(c+a)/(a+b+c)]
=> [a/(a+b)] + [b/(b+c)] + [c/(c+a)] < [2.(a+b+c)]/(a+b+c)
=> [a/(a+b)] + [b/(b+c)] + [c/(c+a)] < 2
Vậy .....
: a<b nên a+a < a+b
=> 2a < a+b (1)
c<d nên c+c < c+d
=> 2c < c+d (2)
m<n nên m+m < m+n
=> 2m < m+n (3)
Từ (1); (2) và (3). 2a + 2c +2m < a+b+c+d+m+n
=> 2(a+c+m) < a+b+c+d+m+n
vậy a+c+m/a+b+c+d+m+n <1/2
đúng ko ạ?