Question

Answer 1

Câu 4:

Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(\Leftrightarrow P=\dfrac{-abc}{abc}=-1\)

Với \(a+b+c\ne0\)

\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b+c}{a+b+c}=1\\ \Leftrightarrow\left\{{}\begin{matrix}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{2a\cdot2b\cdot2c}{abc}=8\)