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\(\Leftrightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Leftrightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\text{Th}1:a+b+c+d=0\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{cases}}\)
\(M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(a+d\right)}=-4\)
\(\text{th}2:a+b+c+d\ne0\Rightarrow a=b=c=d\)
\(\Leftrightarrow M=1+1+1+1=4\)
Vậy....
p/s: đầu tiên nhớ ghi lại cái đề nha :))
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\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
\(M=1+1+1+1=4\)
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Theo đầu bài ta có:
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Do \(a+b+c=259\Rightarrow\hept{\begin{cases}a=259-\left(b+c\right)\\b=259-\left(a+c\right)\\c=259-\left(a+b\right)\end{cases}}\)
Từ đó suy ra:
\(\Leftrightarrow Q=\frac{259-\left(b+c\right)}{b+c}+\frac{259-\left(a+c\right)}{a+c}+\frac{259-\left(a+b\right)}{a+b}\)
\(\Leftrightarrow Q=\left(\frac{259}{b+c}-\frac{b+c}{b+c}\right)+\left(\frac{259}{a+c}-\frac{a+c}{a+c}\right)+\left(\frac{259}{a+b}-\frac{a+b}{a+b}\right)\)
\(\Leftrightarrow Q=\left(259\cdot\frac{1}{b+c}+259\cdot\frac{1}{a+c}+259\cdot\frac{1}{a+b}\right)-\left(\frac{b+c}{b+c}+\frac{a+c}{a+c}+\frac{a+b}{a+b}\right)\)
\(\Leftrightarrow Q=259\cdot\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-\left(1+1+1\right)\)
Do \(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}=15\) nên:
\(\Leftrightarrow Q=259\cdot15-3\)
\(\Leftrightarrow Q=3882\)
a=259-(b+c)
b=259-(c+a)
c=259-(a+b)
Thay vào Q
Q=259-(a+b)/a+b+259-(b+c)/b+c+259-(c+a)/c+a
Q=259/a+b+259/b+c+259/c+a-3
Q=259.(1/a+b+1/c+a+1/b)+c-3
Q=259x15-3
Q=3882
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\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\) => \(\frac{a+b}{c}-1=\frac{a+c}{b}-1=\frac{b+c}{a}-1\)
=> \(\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=\frac{\left(a+b\right)+\left(a+c\right)+\left(b+c\right)}{c+b+a}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
M = \(\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}\) = 2.2.2 = 8
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Ta có: \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
Vậy \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{1}{2}\)
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Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{\left(b+c\right)+\left(c+a\right)+\left(a+b\right)}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
Vậy các tỉ số đó bằng 1/2
Đặt \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=k\)
\(\Rightarrow a=kb+kc\)và \(b=kc+ka\)và \(c=ka+kb\)
\(\Rightarrow a+b+c=kb+kc+kc+ka+ka+kb\)
\(a+b+c=k\left(b+c+c+a+a+b\right)\)
\(a+b+c=k\left[2\left(a+b+c\right)\right]\)
\(\frac{a+b+c}{2\left(a+b+c\right)}=k\)
\(\Rightarrow k=\frac{1}{2}\)
Mà \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=k\)
Vậy \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{1}{2}\)