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Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}2a=b+c\\2b=a+c\\2c=a+b\end{cases}}\)
Vậy \(P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+b}=2\)
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Vì \(a,b,c\ne0\)
\(\Rightarrow\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=2\)
\(\Rightarrow P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=2+2+2=6\)
Ta có : \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
=> \(\frac{a}{b+c}+1=\frac{b}{a+c}+1=\frac{c}{a+b}+1\)
=> \(\frac{a+b+c}{b+c}=\frac{a+b+c}{a+c}=\frac{a+b+c}{a+b}\)
Nếu a + b + c = 0
=> a + b = - c
=> b + c = - a
=> a + c = - b
Khi đó P = \(\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-1+\left(-1\right)+\left(-1\right)=-3\)
Nếu a + b + c \(\ne0\)
=> \(\frac{1}{b+c}=\frac{1}{a+c}=\frac{1}{a+b}\)
=> b + c = a + c = a + b
=> \(\hept{\begin{cases}b+c=a+c\\b+c=a+b\end{cases}\Rightarrow\hept{\begin{cases}a=b\\a=c\end{cases}}\Rightarrow a=b=c}\)
Khi đó P = \(\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}=2+2+2=6\)
=> P = 6
Vậy khi a + b + c = 0 => P = -3
khi a + b + c \(\ne0\) => P = 6
Ta có
a/b+c=b/c+a=c/b+a => a/ b+c +1=b/c+a +1=c/b+a +1
=> a+b+c/b+c=a+b+c/c+a=a+b+c/b+a
=> b+c=c+a=b+a
=> a=b=c
=> B= 2a/a+2b/b+2c/c =2+2+2=6 ( tick nhe
Ta có:\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}\)
*Nếu a+b+c=0
=> a=-(b+c)
b=-(a+c)
c=-(a+b)
Thay 3 ý trên vào P, ta có:
\(P=\frac{b+c}{-\left(b+c\right)}+\frac{a+c}{-\left(a+c\right)}+\frac{a+b}{-\left(a+b\right)}\)
P=-1+(-1)+(-1)
P=-3
Nếu a+b+c khác 0
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{1}{2}\)
\(\frac{a}{b+c}=\frac{1}{2}\) => 2a=b+c (1)
\(\frac{b}{a+c}=\frac{1}{2}\) => 2b=a+c (2)
\(\frac{c}{a+b}=\frac{1}{2}\) => 2c=a+b (3)
(1)-(2)
2a-2b=b-a
3a=3b
=>a=b (4)
(2)-(3)
2b-2c=c-b
3b=3c
=>b=c (5)
Từ (4) và (5)=> a=b=c (mâu thuẫn với đề bài)
Vậy M=-3
Ta có:
a/b+c =b/a+c =c/a+b hay b+c/a =a+c/b =a+b/c =(b+c)+(a+c)+(a+b)a+b+c =2a+2b+2c/a+b+c =2(a+b+c)/a+b+c =2
=>b+c/a =2;a+c/b =2;a+b/c =2
=>P=b+c/a +a+c/b +a+b/c =2+2+2=6
Vậy P=6
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)\(\Rightarrow b+c=2a\)
\(\Rightarrow a+c=2b\)
\(\Rightarrow a+b=2c\)
\(D=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)
\(D=\frac{2a}{a}=\frac{2b}{b}=\frac{2c}{c}\)
\(D=2+2+2\)
\(D=6\)
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
=>b+c=2a
=>a+c=2b
=>a+b=2c
\(D=\frac{b+c}{a}+\frac{a+c}{b}=\frac{a+b}{c}\)
\(D=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}\)
\(D=2+2+2\)
D=6
Vậy D=6
^...^ 
^_^
Vì \(a,b,c\ne0\) nên:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}b+c=2a\\a+c=2b\\a+b=2c\end{cases}}\)
\(\Rightarrow D=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}=2+2+2=6\)
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow b+c=2a\)
\(\Rightarrow a+c=2b\)
\(\Rightarrow a+b=2c\)
\(D=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)
\(D=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}\)
\(D=2+2+2\)
\(D=6\)
Vậy \(D=6\)
Ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Leftrightarrow\)
\(\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{b+c+a+c+a+b}{a+b+c}=2\)
\(\Rightarrow P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=3.2=6\)
bài này có 2 trường hợp nhé =))
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Rightarrow1+\frac{a}{b+c}=1+\frac{b}{a+c}=1+\frac{c}{a+b}\)
\(\Rightarrow\frac{a+b+c}{b+c}=\frac{a+b+c}{a+c}=\frac{a+b+c}{a+b}\)
\(TH1:a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}\Rightarrow P=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-3}\)
\(TH2:a+b+c\ne0\)
\(\Rightarrow\hept{\begin{cases}b+c=a+c\Rightarrow a=b\\a+c=a+b\Rightarrow c=b\\a+b=b+c\Rightarrow a=c\end{cases}\Rightarrow a=b=c}\)
\(\Rightarrow P=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}=2.3=6\)
Vậy P=-3 hay P=6