B

B 

ta có a+b+c=0       =>     a=-b-c,         b=-a-c,            c=-a-b

thay vào A ta được 

 A=(1-(b+c)/b)(1-(a+c)/c)(1-(a+b)/a)

   =(1-1-c/b)(1-1-a/c)(1-1-b/a)

   =(-c/b)(-a/c)(-b/a)

   =(-abc)/abc

    =-1

bạn Nguyễn Thị Lan Hương làm đúng rồi, mk lm cách khác nhé:

           BÀI LÀM

          \(a+b+c=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

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

    \(=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\)

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

\(a+b+c=0\Rightarrow a+b=-c;a+c=-b;b+c=-a\)

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

\(=-\frac{c}{b}\cdot-\frac{a}{c}\cdot-\frac{b}{a}=\frac{-c\cdot-a\cdot-b}{b\cdot c\cdot a}=-1\cdot-1\cdot-1=-1\)

a: Sửa đề: \(\frac{a}{b}=\frac{a+c}{b+d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a}{b}=\frac{bk}{b}=k\)

\(\frac{a+c}{b+d}=\frac{bk+dk}{b+d}=\frac{k\left(b+d\right)}{b+d}=k\)

Do đó: \(\frac{a}{b}=\frac{a+c}{b+d}\)

b: \(\frac{a-b}{a}=\frac{bk-b}{bk}=\frac{b\left(k-1\right)}{bk}=\frac{k-1}{k}\)

\(\frac{c-d}{c}=\frac{dk-d}{dk}=\frac{d\left(k-1\right)}{dk}=\frac{k-1}{k}\)

Do đó: \(\frac{a-b}{a}=\frac{c-d}{c}\)