Question

cho abc khác 0 và \(\dfrac{a-b+c}{c}=\dfrac{b+c-a}{a}=\dfrac{a+c-b}{b}\) Tính P\(=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\) giúp mik nha! help me =='

Answer 1

Ta có: \(a-b-c=0\)

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

Ta thay: \(a=b+c;b=a-c;c=a-b\) vào biểu thức \(A\), ta đc:

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

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

\(\dfrac{b}{a}.\dfrac{-c}{b}.\dfrac{a}{c}=-1\)

Chúc bạn học tốt!