Question

Cho \(\frac{x+2}{2}=\frac{y+3}{3}=\frac{z+4}{4};2x+y+z=11\). Tính x.y.z

Answer 1

Ta có : \(\frac{x+2}{2}=\frac{y+3}{3}=\frac{z+4}{4}=>\frac{2x+4}{4}=\frac{y+3}{3}=\frac{z+4}{4}\)\(=\frac{2x+4+y+3+z+4}{4+3+4}=\frac{2x+y+z+11}{11}=\frac{11+11}{11}=2\)

+)\(\frac{x+2}{2}=2=>x+2=4=>x=2\)

+)\(\frac{y+3}{3}=2=>y+3=6=>y=3\)

+)\(\frac{z+4}{4}=2=>z+4=8=>z=4\)

Vậy x = 2 ; y = 3 ; z = 4

Answer 2

\(\frac{x+2}{2}=\frac{y+3}{3}=\frac{z+4}{4}\Rightarrow\frac{x+2}{2}=\frac{x+2}{2}=\frac{y+3}{3}=\frac{z+4}{4}=\frac{x+2+x+2+y+3+z+4}{2+2+3+4}\)

\(=\frac{2x+y+x+2+2+3+4}{2+2+3+4}=\frac{11+11}{11}=\frac{22}{11}=2\)

\(\Rightarrow\frac{x+2}{2}=2\Rightarrow x=2\)

\(\Rightarrow\frac{y+3}{3}=2\Rightarrow y=3\)

\(\Rightarrow\frac{z+4}{4}=2\Rightarrow z=4\)