Cho $\frac{a+b+c}{b}$= $\frac{b+c-a}{c}$= $\frac{a-c+b}{a}$ Tính: A= $\frac{(a+c)(c-b)(b-a)}{abc}$

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1. Đáp án: $A=8$ hoặc $A=-1$

   Giải thích các bước giải:

   Ta có:

   $\dfrac{a+b+c}{b}=\dfrac{b+c-a}{c}=\dfrac{a-c+b}{a}$

   $\to 1+\dfrac{a+c}{b}=1+\dfrac{b-a}{c}=1+\dfrac{-c+b}{a}$

   $\to \dfrac{a+c}{b}=\dfrac{b-a}{c}=\dfrac{-c+b}{a}$

   $\to A=\dfrac{(a+c)(c-b)(b-a)}{abc}=-\dfrac{a+c}{b}.\dfrac{-c+b}{a}.\dfrac{b-a}{c}=-(\dfrac{a+c}{b})^3$

   Lại có:

   $\dfrac{a+b+c}{b}=\dfrac{b+c-a}{c}=\dfrac{a-c+b}{a}=\dfrac{b+c-a+a-c+b}{c+a}=\dfrac{2b}{c+a}$

   $\to (a+b+c)(a+c)=2b^2$

   $\to (a+c)^2+b(a+c)=2b^2$

   $\to (a+c)^2+b(a+c)-2b^2=0$

   $\to (a+c)^2+2b(a+c)-b(a+c)-2b^2=0$

   $\to (a+c+2b)(a+c)-b(a+c+2b)=0$

   $\to (a+c+2b)(a+c-b)=0$

   $\to a+c+2b=0$

   $\to a+c=-2b$

   $\to \dfrac{a+c}{b}=-2$

   $\to A=-(-2)^3=8$

   Hoặc $a+c-b=0\to a+c=b\to \dfrac{a+c}{b}=1$

   $\to A=-1^3=-1$

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