Question

cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)  a+b+c \(\ne\) 0 biết a=2003 . Tính b,c

Answer 1

\(a+b+c\ne0\) biết a = 2003

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

\(\frac{a}{b}=\frac{c}{a}\Rightarrow bc=a^2=2003^2\)

\(\Rightarrow b=2003;c=2003\\\)

Vậy b = 2003;c = 2003

Answer 2

ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

\(\begin{cases}=>a=b\\=>b=c\\=>c=a\end{cases}=>a=b=c}\)

\(b=2003;c=2003\)

Answer 3

taco:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

\(=>a=b\)

\(=>b=c\)

\(=>c=a\)

\(< =>a=b=c\)

\(=>b=2003;c=2003\)

vậy :\(b=2003;c=2003\)