Áp dụng t/c của dãy tỉ số bằng nhau có:
`[a+b-c]/c=[b+c-a]/a=[c+a-b]/b=[a+b-c+b+c-a+c+a-b]/[a+b+c]=[a+b+c]/[a+b+c]=1`
`@[a+b-c]/c=1=>a+b-c=c=>a+b=2c`
`@[b+c-a]/a=1=>b+c=2a`
`@[c+a-b]/b=1=>c+a=2b`
Có: `P=(1+b/a)(1+c/b)(1+a/c)`
`P=[a+b]/a . [b+c]/b . [a+c]/c`
`P=[2c]/a . [2a]/b . [2b]/c`
`P=8`
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{\left(a+b-c\right)+\left(b+c-a\right)+\left(c+a-b\right)}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
Do đó: \(P=\left(1+\dfrac{b}{a}\right).\left(1+\dfrac{c}{b}\right).\left(1+\dfrac{a}{c}\right)=\dfrac{a+b}{a}.\dfrac{b+c}{b}.\dfrac{c+a}{c}=\dfrac{2c}{a}.\dfrac{2a}{b}.\dfrac{2b}{c}=\dfrac{8abc}{abc}=8\)
( abc≠0 ko thể khẳng định được a+b+c≠0 ví dụ a=2, b=-1; c=-1)
Xét 2 TH:
TH1: a+b+c=0 ⇒ a+b=-c; b+c=-a; c+a=-b\(\Rightarrow A=\left(\dfrac{a+b}{a}\right)\cdot\left(\dfrac{b+c}{b}\right)\cdot\left(\dfrac{c+a}{c}\right)=\left(-\dfrac{c}{a}\right)\cdot\left(-\dfrac{a}{b}\right)\cdot\left(\dfrac{-b}{c}\right)=-1\)
TH2: a+b+c ≠ 0
Theo tính chất của dãy tỉ số bằng nhau ta có:
\(\Rightarrow\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b+c}{a+b+c}=1\Rightarrow a+b=2c;b+c=2a;c+a=2b\)
\(\Rightarrow P=\left(\dfrac{a+b}{a}\right)\cdot\left(\dfrac{b+c}{b}\right)\cdot\left(\dfrac{c+a}{c}\right)=\dfrac{2c}{a}\cdot\dfrac{2a}{b}\cdot\dfrac{2b}{c}=8\)
