a)\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{5}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{z^2}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{z^2}{25}=\dfrac{x^2-y^2+z^2}{9-49+25}=\dfrac{-60}{-15}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4.9=36\\y^2=4.49=196\\z^2=4.25=100\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm14\\z=\pm10\end{matrix}\right.\)
b)
\(\left\{{}\begin{matrix}5x=2y\\3y=5z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{5}\\\dfrac{y}{5}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y+z}{2+5+3}=\dfrac{-60}{10}=-6\)
\(\Rightarrow\left\{{}\begin{matrix}x=-6.2=-12\\y=-6.5=-30\\z=-6.3=-18\end{matrix}\right.\)
a. \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{5}\) và \(x^2-y^2+z^2=-60\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{5}=\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{z^2}{25}=\dfrac{x^2-y^2+z^2}{9-49+25}=\dfrac{-60}{-15}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=4\Rightarrow x=4.3=12\\\dfrac{y}{7}=4\Rightarrow y=4.7=28\\\dfrac{z}{5}=4\Rightarrow z=5.4=20\end{matrix}\right.\)
Vậy...........
b. Đề bài câu b chắc là sai . mk sửa lại
\(5x=2y;3y=5z\) và \(x+y+z=-60\)
Theo đề bài ta có :
\(\begin{matrix}5x=2y\\3y=5z\end{matrix}\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{5}\\\dfrac{y}{5}=\dfrac{z}{3}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{3}\)
Áp dụng t/c dãy tỉ số bằng nhau :
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y+z}{2+5+3}=\dfrac{-60}{10}=-6\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-6\Rightarrow x=2.\left(-6\right)=-12\\\dfrac{y}{5}=-6\Rightarrow y=5.\left(-6\right)=-30\\\dfrac{z}{3}=-6\Rightarrow z=3.\left(-6\right)=-18\end{matrix}\right.\)
Vậy..........
