Question

tìm x,y,z

\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}\)và \(x+y+z=48\)

Answer 1

Theo t/c của dãy ỉ số bằng nhau ta có  :

\(=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)

\(\Rightarrow\frac{12x-15y}{7}=0\Rightarrow12x-15y=0\Rightarrow12x=15y=\frac{x}{15}=\frac{y}{12}\)

\(\frac{15y-20z}{11}=0\Rightarrow15y-20z=0\Rightarrow15y=20z=\frac{y}{20}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{y}{20}=\frac{z}{15}\)

\(=\frac{x}{75}=\frac{y}{60}=\frac{z}{45}=\frac{x+y+z}{75+60+45}=\frac{48}{180}=\frac{4}{15}\)

\(\Rightarrow x=75.\frac{4}{15}=20\)

\(\Rightarrow y=60.\frac{4}{15}=16\)

\(\Rightarrow z=45.\frac{4}{15}=12\)

Vậy x , y ,z = 20 ; 16 ; 12 

Answer 2

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{12x-15y}{7}=\frac{20z-12x}{8}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)

Suy ra : \(\frac{12x-15y}{7}=0\Rightarrow12x-15y=0\Rightarrow12x=15y\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(\frac{15y-20z}{11}=0\Rightarrow15y-20z=0\Rightarrow15y=20z\Rightarrow\frac{y}{20}=\frac{z}{15}\)

Suy ra :\(\frac{x}{75}=\frac{y}{60}=\frac{z}{45}=\frac{x+y+z}{75+60+45}=\frac{48}{180}=\frac{4}{15}\)

\(\Rightarrow x=75.\frac{4}{15}=20;y=60.\frac{4}{15}=16;z=45.\frac{4}{15}=12\)