Question

Cho các số a,b,c thỏa mãn: \(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}.\) Tính P= \(\left(\dfrac{a}{b}+1\right)\left(\dfrac{b}{c}+1\right)\left(\dfrac{c}{a}+1\right)\)

Answer 1

+) Nếu \(a+b+c=0\)

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

\(\Leftrightarrow P=\left(\dfrac{a}{b}+1\right)\left(\dfrac{b}{c}+1\right)\left(\dfrac{c}{a}+1\right)\)

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

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

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

\(=-1\)

+) Nếu \(a+b+c\ne0\)

Theo t.c dãy tỉ số bằng nhau ta có

\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)

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