Question

Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\); \(a+b+c\ne0\);\(a=2003\)

Tính b và c

Answer 1

ta có:

a/b = b/c = c/a = [a+b+c]/[a+b+c] = 1 [ap dung tc day ti so bg nhau] 

=> a/b = b/c = c/a = 1 =>a =b=c=2003

Answer 2

\(\frac{a}{b}=\frac{2003}{b}=\frac{b}{c}\)

\(=>2003.c=b^2\)

\(=>2003.c.b=b^3\)

\(=>2003.b=\frac{b^3}{c}\)

\(\frac{b}{c}=\frac{c}{a}=\frac{c}{2003}\)

\(=>2003.b=c^2=\frac{b^3}{c}\)

\(=>c^2=\frac{b^3}{c}\)

\(=>c^3=b^3\)

\(=>c=b\)

Lại có:\(\frac{a}{b}=\frac{c}{a}=>\frac{2003}{b}=\frac{c}{2003}\)

\(=>2003.2003=b.c\)

\(=>2003^2=b.b=b^2\)

\(=>b=2003=c\)

Vậy b=c=2003