Ta có:\(\dfrac{b+c-3a}{a}=\dfrac{a+c-3b}{b}=\dfrac{a+c-3c}{c}\)
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\)
Nếu a+b+c=0
\(\Rightarrow\)\(\left\{{}\begin{matrix}b+c=a\\a+c=b\\a+b=c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow M=-1\)
Nếu a+b+c\(\ne\)0
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\)
\(\Rightarrow M=8\)
Tính \(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\left(1\right)\)
Ta có :
\(\dfrac{b+c-3a}{a}=\dfrac{a+c-3b}{b}=\dfrac{a+b-3c}{c}\)
\(\Rightarrow\dfrac{b+c-3a}{a}+3=\dfrac{a+c-3b}{b}+3=\dfrac{a+b-3c}{c}+3\)
\(\Rightarrow\dfrac{b+c-3a}{a}+\dfrac{3a}{a}=\dfrac{a+c-3b}{b}+\dfrac{3b}{b}=\dfrac{a+b-3c}{c}+\dfrac{3c}{c}\)
\(\Rightarrow\dfrac{b+c-3a+3a}{a}=\dfrac{a+c-3b+3b}{b}=\dfrac{a+b-3c+3c}{c}\)
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta được :
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{2a+2b+2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)Suy ra :
\(\left\{{}\begin{matrix}\dfrac{b+c}{a}=2\\\dfrac{a+c}{b}=2\\\dfrac{a+b}{c}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{b+c}{2}=a\left(2\right)\\\dfrac{a+c}{2}=b\left(3\right)\\\dfrac{a+b}{2}=c\left(4\right)\end{matrix}\right.\)
THAY (2),(3),(4)vào(1) ta được :
M=\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\dfrac{b+c}{2}.\dfrac{a+c}{2}.\dfrac{a+b}{2}}\)
\(=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\dfrac{\left(b+c\right)\left(a+c\right)\left(a+b\right)}{2.2.2}}=2.2.2=2^3=8\)
Chúc bạn thành công trên con đường học tập
Mình bổ sung
Mặt khác nếu a+b+c=0 thì
\(\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{b+c+a+c+a+b}{a+b+c}=0\)
ta chia ra 3 trường hợp sau
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b+c}{a}=0\\\dfrac{a+c}{b}=0\\\dfrac{a+b}{c}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b+c=a\\a+c=b\\a+b=c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c=a-b\\a=b-c\\b=c-a\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}c=-\left(a+b\right)\\a=-\left(b+c\right)\\b=-\left(c+a\right)\end{matrix}\right.\)
\(\)Ta thay thế vào M được
\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{a+b+c}=\)\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{-\left(b+c\right).-\left(c+a\right).-\left(a+b\right)}=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{-[\left(b+c\right)\left(c+a\right)\left(a+b\right)]}=-1\)
VẬY M có hai giá trị là 8 và -1
