Question

Biết \(\frac{a}{2}=\frac{b}{3};\frac{a}{4}=\frac{c}{9}\) và \(^{a^3+b^3+c^3=-1009}\)

Answer 1

\(\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{4}=\frac{b}{6}=\frac{c}{9}\)

\(\Rightarrow\frac{a^3}{4^3}=\frac{b^3}{6^3}=\frac{c^3}{9^3}=\frac{a^3+b^3+c^3}{64+216+729}=\frac{-1009}{1009}=-1\)

=>a³=-64=>a=-4

b³=216=>b=-6

c³=-729=>c=-9

Vậy (a;b;c)=(-4;-6;-9)

Answer 2

\(\frac{a}{2}=\frac{b}{3};\frac{a}{4}=\frac{c}{9}\)

suy ra: \(\frac{b}{12}=\frac{a}{8}=\frac{c}{18}suyra\frac{b^3}{1728}=\frac{a^3}{512}=\frac{c^3}{5832}\)

suy ra \(\frac{b^3+a^3+c^3}{1728+512+5832}=\frac{-1009}{8072}=\frac{-1}{8}\)

 a/8= -1/8 suy ra a=-1

b/12=-1/8 suy ra b= -3/2

c/18=-1/8 suy ra c = -9/4

b/