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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
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{a}{a+b}\cdot b=\frac{c}{b+c}\cdot b\)
\(\Rightarrow\frac{a}{a+b}=\frac{c}{b+c}\Rightarrow a\left(b+c\right)=c\left(a+b\right)\Rightarrow ab+ac=ac+bc\Rightarrow ab=bc\Rightarrow a=c\left(1\right)\)
\(\frac{ab}{a+b}=\frac{ac}{a+c}=\frac{b}{a+b}\cdot a=\frac{c}{a+c}\cdot a\)
\(\Rightarrow\frac{b}{a+b}=\frac{c}{a+c}\Rightarrow b\left(a+c\right)=c\left(a+b\right)\Rightarrow ab+bc=ac+bc\Rightarrow ab=ac\Rightarrow b=c\left(2\right)\)
\(\frac{bc}{b+c}=\frac{ac}{a+c}=\frac{b}{b+c}\cdot c=\frac{a}{a+c}\cdot c\)
\(\Rightarrow\frac{b}{b+c}=\frac{a}{a+c}\Rightarrow b\left(a+c\right)=a\left(b+c\right)\Rightarrow ab+bc=ab+ac\Rightarrow bc=ac\Rightarrow a=b\left(3\right)\)
từ \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow a=b=c\)
\(\Rightarrow M=\frac{ab+bc+ac}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
Ta có:\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\iff\)\(\frac{abc}{ac+bc}=\frac{abc}{ab+ac}=\frac{abc}{bc+ba}\)
\(\iff\) \(ac+bc=ab+ac=bc+ba\)
+)\(ac+bc=ab+ac\)
\(\implies\)\(bc=ab\)
\(\implies\) \(c=a\left(1\right)\)
+)\(ab+ac=bc+ba\)
\(\implies\) \(ac=bc\)
\(\implies\) \(a=b\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\)
\(\implies\) \(a=b=c\)
\(\implies\) \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{aa+bb+cc}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
Vậy \(M=1\)
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}\)
+) a+b+c=0 => \(\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\Rightarrow P=-3\)
+) a+b+c khác 0 => \(\hept{\begin{cases}a=\frac{1}{2}\left(b+c\right)\\b=\frac{1}{2}\left(a+c\right)\\c=\frac{1}{2}\left(b+a\right)\end{cases}}\)
\(\Rightarrow P=\frac{3}{2}\)
Vậy: P = 3/2 hoac P=-3
Ta có: \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
Suy ra:
\(\frac{a}{b+c}=\frac{1}{2}\Rightarrow a=\frac{b+c}{2}=\frac{1}{2}\times\left(b+c\right)\)
\(\frac{b}{a+c}=\frac{1}{2}\Rightarrow b=\frac{a+c}{2}=\frac{1}{2}\times\left(a+c\right)\)
\(\frac{c}{a+b}=\frac{1}{2}\Rightarrow c=\frac{a+b}{2}=\frac{1}{2}\times\left(a+b\right)\)
Thay \(a=\frac{1}{2}\times\left(b+c\right)\); \(b=\frac{1}{2}\times\left(a+c\right)\); \(c=\frac{1}{2}\times\left(a+b\right)\) vào P ta được:
\(\frac{b+c}{\frac{1}{2}\times\left(b+c\right)}+\frac{c+a}{\frac{1}{2}\times\left(a+c\right)}+\frac{a+b}{\frac{1}{2}\times\left(a+b\right)}\)
\(=\frac{\text{ }1\text{ }}{\frac{1}{2}}+\frac{1}{\frac{1}{2}}+\frac{1}{\frac{1}{2}}\)
\(=2+2+2=6\)
Vậy giá trị của P là 6