Question

Tìm x , y , z thỏa mãn : \(\dfrac{3x+5}{15}=\dfrac{3y+4}{12}=\dfrac{z+1}{3}\) và x² + y² + z² = 200

Answer 1

\(\dfrac{3x+5}{15}=\dfrac{3y+4}{12}=\dfrac{z+1}{3}\Rightarrow\dfrac{3x}{15}+\dfrac{5}{15}=\dfrac{3y}{12}+\dfrac{4}{12}=\dfrac{z}{3}+\dfrac{1}{3}\)

\(\Rightarrow\dfrac{x}{5}+\dfrac{1}{3}=\dfrac{y}{4}+\dfrac{1}{3}=\dfrac{z}{3}+\dfrac{1}{3}\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{25+16+9}=\dfrac{200}{50}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=25.4=100\\y^2=16.4=64\\z^2=9.4=36\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\y=8\\z=6\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-10\\y=-8\\z=-6\end{matrix}\right.\)