Question

Tìm x,y,z biết: \(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\) và xy= 1200

Answer 1

Ta có: \(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\)

\(\Rightarrow\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\)

\(\Rightarrow\frac{x}{15}-\frac{9}{15}=\frac{y}{20}-\frac{12}{20}=\frac{z}{40}-\frac{24}{40}\)

\(=\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\)

\(\Rightarrow\frac{x^2}{15^2}=\frac{y^2}{20^2}=\frac{z^2}{40^2}=\frac{xy}{15.20}=\frac{1200}{300}=4=2^2\)

\(\Rightarrow\begin{cases}x^2=2^2.15^2=30^2\\y^2=2^2.20^2=40^2\\z^2=2^2.40^2=80^2\end{cases}\)\(\Rightarrow\begin{cases}x\in\left\{30;-30\right\}\\y\in\left\{40;-40\right\}\\z\in\left\{80;-80\right\}\end{cases}\)

Vậy giá trị (x;y;z) tương ứng thỏa mãn là: (30;40;80) ; (-30;-40;-80)