Question

Cho a,b,c là các số khác 0 sao cho

\(\dfrac{a+b-c}{c}\)= \(\dfrac{a-b+c}{b}\)= \(\dfrac{-a+b+c}{a}\)

Tính giá trị biểu thức

M=\(\dfrac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}\)

Answer 1

Lời giải:

TH1: $a+b+c=0$

Khi đó: \(a+b=-c; b+c=-a; c+a=-b\)

\(\Rightarrow M=\frac{(-c)(-a)(-b)}{abc}=\frac{-abc}{abc}=-1\)

TH2: \(a+b+c\neq 0\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=\frac{a+b-c+a-b+c-a+b+c}{c+b+a}=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow \left\{\begin{matrix} a+b-c=c\\ a-b+c=b\\ -a+b+c=a\end{matrix}\right.\Rightarrow a+b=2c; a+c=2b; b+c=2a\)

\(\Rightarrow M=\frac{2c.2a.2b}{abc}=\frac{8abc}{abc}=8\)

Answer 2

Ta có:

\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\)

\(=\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}\)

\(=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-c}{c}=1\\\dfrac{a-b+c}{b}=1\\\dfrac{-a+b+c}{a}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)(1)

Thay (1) vào M ta được

\(M=\dfrac{2c.2a.2b}{abc}=\dfrac{8abc}{abc}=8\)