Question

Cho a,b,c khác 0. a+b+c=0. tính \(\frac{\left(a+b\right)\left(b+c\right)\left(c+b\right)}{abc}\)

Answer 1

a+b+c = 0 =>a+b=-c ; b+c=-a ; c+a=-b

=> (a+b).(b+c).(c+a)/abc = (-c).(-a).(-b)/abc = -abc/abc = -1

k mk nha

Answer 2

Ta có : a + b + c  = 0 

=> a + b = -c

=> b + c = -a

=> c + a = -b

Vậy \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)

Answer 3

Ta có: a+b+c=0 

=> a + b = -c 

     b + c = -a

     c + a = -b

\(\Rightarrow\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-\frac{abc}{abc}=-1\)

Vậy \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=1\)