Question

[Nguyễn Võ Hoàng Minh Hạnh]

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

Answer 1

[Hoang Hung Quan]

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

\(\Rightarrow\left\{\begin{matrix}\frac{12x-15y}{7}=0\Rightarrow12x-15y=0\Rightarrow\frac{x}{15}=\frac{y}{20}\\\frac{15y-20z}{11}=0\Rightarrow15y-20z=0\Rightarrow\frac{y}{20}=\frac{z}{15}\\\frac{20z-12x}{9}=0\Rightarrow20z-12x=0\Rightarrow\frac{z}{20}=\frac{x}{12}\end{matrix}\right.\)

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

\(\Rightarrow\left\{\begin{matrix}x=75.\frac{4}{15}=20\\y=60.\frac{4}{15}=16\\z=45.\frac{4}{15}=12\end{matrix}\right.\)

Vậy: \(\left\{\begin{matrix}x=20\\y=16\\z=12\end{matrix}\right.\)

Answer 2

[Nguyễn Hoàng Thiên Ân]

75 ở đâu ra vậy ạ??