\(A=\left(\frac{a+b}{b}\right).\left(\frac{b+c}{c}\right).\left(\frac{c+a}{a}\right)\)
Vì \(a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\c+b=-a\end{cases}}\)
\(\Rightarrow A=\frac{-c}{b}.\left(\frac{-a}{c}\right).\left(\frac{-b}{a}\right)=-1\)
Ta có: \(a+b+c=0\)
\(\Rightarrow b+a=-c\)
\(\Rightarrow c+b=-a\)
\(\Rightarrow a+c=-b\)
Ta có: \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(\Rightarrow A=\left(\frac{b+a}{b}\right)\left(\frac{c+b}{c}\right)\left(\frac{a+c}{a}\right)\)
\(\Rightarrow A=\left(\frac{-c}{b}\right)\left(\frac{-a}{c}\right)\left(\frac{-b}{a}\right)\)
\(\Rightarrow A=-1\)
~~k cho mik nha~~
Ta có: a+b+c=0a+b+c=0
\Rightarrow b+a=-c⇒b+a=−c
\Rightarrow c+b=-a⇒c+b=−a
\Rightarrow a+c=-b⇒a+c=−b
Ta có: A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)A=(1+
b
a
)(1+
c
b
)(1+
a
c
)
\Rightarrow A=\left(\frac{b+a}{b}\right)\left(\frac{c+b}{c}\right)\left(\frac{a+c}{a}\right)⇒A=(
b
b+a
)(
c
c+b
)(
a
a+c
)
\Rightarrow A=\left(\frac{-c}{b}\right)\left(\frac{-a}{c}\right)\left(\frac{-b}{a}\right)⇒A=(
b
−c
)(
c
−a
)(
a
−b
)
\Rightarrow A=-1⇒A=−1