\(\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\)
\(\Rightarrow\dfrac{x-9}{15}=\dfrac{y-12}{20}=\dfrac{z-24}{40}=k\)
\(\Rightarrow\left(15k+9\right)\left(20k+12\right)=1200\)
\(\Rightarrow3.4\left(5k+3\right)\left(5k+3\right)=1200\)
\(\Rightarrow\left(5k+3\right)\left(5k+3\right)=1200:3:4\)
\(\Rightarrow\left(5k+3\right)^2=100\)
\(\Rightarrow\left[{}\begin{matrix}5k+3=10\\5k+3=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}k=\dfrac{7}{5}\\k=-\dfrac{13}{5}\end{matrix}\right.\)
+) Với \(k=\dfrac{7}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{5}.15+9\\y=\dfrac{7}{5}.20+12\\z=\dfrac{7}{5}.40+24\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=80\end{matrix}\right.\)
+) Với \(k=-\dfrac{13}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{13}{5}.15+9\\y=-\dfrac{13}{5}.20+12\\z=-\dfrac{13}{5}.40+24\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-30\\y=-40\\z=-80\end{matrix}\right.\)
Vậy...................................
Ta có: (dfrac{15}{x-9}=dfrac{20}{y-12}=dfrac{40}{z-24})
(Rightarrowdfrac{x-9}{15}=dfrac{y-12}{20}=dfrac{z-24}{40})
(Rightarrowdfrac{x}{15}-dfrac{3}{5}=dfrac{y}{20}-dfrac{3}{5}=dfrac{z}{40}-dfrac{3}{5})
(Rightarrowdfrac{x}{15}=dfrac{y}{20}=dfrac{z}{40})
Đặt (dfrac{x}{15}=dfrac{y}{20}=dfrac{z}{40}=k)
(Rightarrowleft[{}egin{matrix}x=15k\y=20k\z=40kend{matrix} ight.) (left(1 ight))
Thay (left(1 ight)) vào đề bài:
(15k.20k=1200)
(Rightarrow300k^2=1200)
(Rightarrow k^2=4Rightarrow k=pm2)
Khi (k=2) thì: (left[{}egin{matrix}x=15.2=30\y=20.2=40\z=40.2=80end{matrix} ight.)
Khi (k=-2) (thì:) (left[{}egin{matrix}x=15left(-2 ight)=-30\y=20left(-2 ight)=-40\z=40left(-2 ight)=-80end{matrix} ight.)
Vậy ta tìm đc 2 cặp giá trị (x,y,z) thỏa mãn đề bài:
(left[{}egin{matrix}x=30\y=40\z=80end{matrix} ight.) và (left[{}egin{matrix}x=-30\y=-40\z=-80end{matrix} ight.).
