Question

cho a,b,c \(\ne\)0 thỏa mãn a+b+c = 0 thỏa mãm 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

ddap an la bang -1 

Answer 2

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

Vì \(a+b+c=0\)

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

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

Answer 3

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~~

Answer 4

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