Question

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

Trả lời : \(\left(x;y;z\right)\)=

Answer 1

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

12x-15y=0\(\Rightarrow4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\)

\(20z-12x=0\Rightarrow20z=12x\Rightarrow5z=3x\Rightarrow\frac{x}{5}=\frac{z}{3}\)

\(15y-20z=0\Rightarrow15y=20z\Rightarrow3y=4z\Rightarrow\frac{y}{4}=\frac{z}{3}\)

suy ra x/5=y/4=z/3=(x+y+z)/(5+4+3)=48/12=4

ta có:x=4x5=20

y=4x4=16

z=4x3=12