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\(\hept{\begin{cases}a< b\Rightarrow2a< a+b\\c< d\Rightarrow2c< c+d\\m< n\Rightarrow2m< m+n\end{cases}}\)
\(\Rightarrow2\left(a+c+m\right)< a+b+c+d+m+n\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\left(đpcm\right)\)
Do a < b < c < d < m < n
=> 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)
a < b \(\Rightarrow\) 2a < a + b
b < d \(\Rightarrow\) 2b < c + d
m < n \(\Rightarrow\) 2m < m + n
\(\Rightarrow\) 2a + 2b + 2m = 2 ( a + b + m ) < ( a + b + c + d + m + n ) . Do đó
a + b + m/a + b + c + d + m + n < 1/2 \(\Rightarrow\) ( đpcm )
Do a<b<c<d<m<n
=>a+c+m<b+d+n
=>2(a+c+m)<a+b+c+d+m+n
=>\(\frac{2\left(a+c+m\right)}{a+b+c+d+m+n}<1\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}<\frac{1}{2}\)
a<b=>2a<a+b
c<d=>2c<c+d
m<n=>2m<m+n
=>2(a+c+m)<a+b+c+d+m+n
=>\(\frac{2\left(a+c+m\right)}{a+b+c+d+m+n}<\frac{a+b+c+d+m+n}{a+b+c+d+m+n}=1\)
<=>\(\frac{a+c+m}{a+b+c+d+m+n}<\frac{1}{2}\)(đpcm)
Cho 6 số nguyên dương thỏa mãn : a<b<c<d<m<n
Chứng minh rằng: \(\frac{a+d}{a+b+c+d+m+n}<\frac{1}{3}\)
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)
Cho 6 số nguyên dương a < b < c < d < m < n
Chứng minh rằng \(\frac{a+c+m}{a+b+c+d+m+n}<\frac{1}{2}\)
a < b \(\Rightarrow\) 2a < a + b ; c < d \(\Rightarrow\) 2c < c + d ; m < n \(\Rightarrow\) 2m < m + n
Suy ra 2a + 2c + 2m = 2(a + c + m) < (a + b + c + d + m + n). Do đó
\(\frac{a+c+m}{a+b+c+d+m+n}<\frac{1}{2}\) (đpcm)
a < b => 2a < a + b ; c < d => 2c < c +d ; m < n =>2m < m + n
Suy ra 2a + 2c + 2m = 2.(a+c+m) < a + b + c + d + m + n. Do đó :
\(\frac{a+c+m}{a+b+c+d+m+n}<\frac{1}{2}\) (đpcm)
\(\frac{a+c+m}{a+b+c+d+m+n}<\frac{a+c+m}{6a}\)
\(\frac{a+c+m}{6a}<\frac{3n}{6a}\)
=> \(\frac{a+c+m}{a+b+c+d+m+n}<\frac{3n}{6a}=\frac{1}{2}.\frac{n}{a}=\frac{1}{2}:\frac{a}{n}\)
Vì a>n nên a/n > 1 => 1/2 : a/n <1/2
Vậy \(\frac{a+c+m}{a+b+c+d+m+n}<\frac{3n}{6a}<\frac{1}{2}\)
Ta có \(\frac{1}{2}=\frac{a+c+m}{a+m+c+a+m+c}=\frac{a+c+m}{2.\left(a+c+m\right)}\)
\(\frac{a+c+m}{a+b+c+d+m+n}=\frac{a+c+m}{a+c+m+d+m+n}\)
Vì a<b;c<d;m<n
=>a+c+m<b+d+n
=2(a+c+m)<a+c+m+b+d+n
=>\(\frac{a+c+m}{2.\left(a+c+m\right)}>\frac{a+c+m}{a+b+c+d+m+n}\)
=>\(\frac{1}{2}>\frac{a+c+m}{a+b+c+d+m+n}\)(ĐPCM)
a < b < c < d < m
=> a + d < c + m + n
=> 3 ( a + d ) < a + b + c + d + m + n
\(\Rightarrow\frac{3\left(a+d\right)}{a+b+c+d+m+n}< 1\)
\(\Rightarrow\frac{a+d}{a+b+c+d+m+n}< \frac{1}{3}\) ( Đpcm )