Question

cho a,b,c thỏa mãn a,b,c#0 và ab+bc+ca=0

Tính P=\(\frac{(a+b)(b+c)(c+a)}{abc}\)

Answer 1

Ta có:

\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(ab+ac+b^2+bc\right)\left(c+a\right)}{abc}\)

\(=\dfrac{b^2\left(c+a\right)}{abc}=\dfrac{b\left(c+a\right)}{ac}=\dfrac{bc+ab}{ac}=\dfrac{-ac}{ac}=-1\)

Answer 2

Ta có: \(ab+bc+ca=0\) => \(bc+ca=-ab\)

Ta lại có: P = \(\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc}\)

= \(\dfrac{\left(a+b\right)\left(ba+bc+ca+c^2\right)}{abc}\) = \(\dfrac{\left(a+b\right)c^2}{abc}\) ( ab + ac+ bc = 0)

= \(\dfrac{ac^2+bc^2}{abc}\) = \(\dfrac{c\left(ac+bc\right)}{abc}\) = \(\dfrac{c.\left(-ab\right)}{abc}\) = \(-1\)
P/s: Bn có thể lm bài này theo 3 cách!