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\(12x=-15y=10z\)
\(\Rightarrow\dfrac{12x}{60}=\dfrac{-15y}{60}=\dfrac{10z}{60}\)
\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{-4}=\dfrac{z}{6}\)
Đặt: \(\dfrac{x}{5}=\dfrac{y}{-4}=\dfrac{z}{6}=k\)
\(\Rightarrow x=5k;y=-4k;z=6k\)
Ta có: \(xyz=120\)
\(\Rightarrow5k\cdot-4k\cdot6k=120\)
\(\Rightarrow-120k^3=120\)
\(\Rightarrow k^3=-1\)
\(\Rightarrow k=-1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-1\\\dfrac{y}{-4}=-1\\\dfrac{z}{6}=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-1\cdot5=-5\\y=-4\cdot-1=4\\z=-1\cdot6=-6\end{matrix}\right.\)
Vậy: ...
\(\frac{12x-5y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20x}{7+9+11}=0\)\(=0\)
=> 12x - 15y =0
=> 12x = 15y
=> \(\frac{x}{15}=\frac{y}{12}\)
=> \(\frac{x}{60}=\frac{y}{48}\)
20z - 12x = 0
=> 20z = 12x
=> \(\frac{x}{20}=\frac{z}{12}\)
=> \(\frac{x}{60}=\frac{z}{36}\)
=> \(\frac{x}{60}=\frac{y}{48}=\frac{z}{36}=\frac{x+y+z}{60+48+36}=\frac{48}{144}=\frac{1}{3}\)
=> x = 1 . 60 : 3 = 20
y = 1 . 48 : 3 = 16
z = 1 . 36 : 3 = 12
Ta có:
\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)
Đề bài gì lạ vậy
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=\frac{0}{27}=0\)
\(\Rightarrow\hept{\begin{cases}12x-15y=0\\15y-20z=0\end{cases}\Rightarrow\hept{\begin{cases}12x=15y\\15y=20z\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{15}=\frac{y}{12}\\\frac{y}{20}=\frac{z}{15}\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{75}=\frac{y}{60}\\\frac{y}{60}=\frac{z}{45}\end{cases}\Rightarrow}\frac{x}{75}=\frac{y}{60}=\frac{z}{45}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{75}=\frac{y}{60}=\frac{z}{45}=\frac{x+y+z}{75+60+45}=\frac{48}{180}=\frac{4}{15}\)
=> x = 75.4 : 15 = 20 ;
y = 60.4 : 15 = 16
z = 45.4 : 15 = 12
Vậy x = 20 ; y = 16 ; z = 12
\(12x=15y=28z\)
\(\Rightarrow\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{15}\)
\(\Rightarrow\dfrac{x^3}{35^3}=\dfrac{y^3}{28^3}=\dfrac{z^3}{15^3}=\dfrac{xyz}{35.28.15}=\dfrac{14700}{14700}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=35\\y=28\\z=15\end{matrix}\right.\)
\(12x=15y=28z\Rightarrow\dfrac{x}{\dfrac{1}{12}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{28}}\)
Đặt \(\dfrac{x}{\dfrac{1}{12}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{28}}=k\Rightarrow x=\dfrac{1}{12}k;y=\dfrac{1}{15}k;z=\dfrac{1}{28}k\)
\(xyz=14700\\ \Rightarrow\dfrac{1}{12}k\cdot\dfrac{1}{15}k\cdot\dfrac{1}{28}k=14700\\ \Rightarrow\dfrac{1}{5040}k^3=14700\\ \Rightarrow k^3=74088000\\ \Rightarrow k=420\\ \Rightarrow\left\{{}\begin{matrix}x=35\\y=28\\z=15\end{matrix}\right.\)