Question

Cho a,b,c khác 0 thoả mãn a+b+c=0 . Tính A=\(\left(1+\frac{a}{b}\right)\)\(\left(1+\frac{b}{c}\right)\)\(\left(1+\frac{c}{a}\right)\)

Answer 1

Ta có: a+b+c=0 => a+b=-c;b+c=-a;a+c=-b

=>\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{b+a}{b}.\frac{c+b}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)

Answer 2

mày đi mà hỏi

Answer 3

Có ai giúp mình bài toán với Cần làm gấp

Answer 4

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

thế vào đề , ta có :

   \(\left(1+\frac{-b-c}{b}\right)\left(1+\frac{-a-c}{c}\right)\left(1+\frac{-a-b}{a}\right)\)^b)

^{<=>  \(\left(\frac{b-b-c}{b}\right)\left(\frac{c-a-c}{c}\right)\left(\frac{a-b-a}{a}\right)\)}

^{<=>  \(\frac{-c}{b}\frac{-a}{c}\frac{-b}{a}\)}

<=> \(\frac{-\left(a\cdot b\cdot c\right)}{a\cdot b\cdot c}\)=  -1

Vậy A= -1