a )\(\dfrac{x}{3}\) = \(\dfrac{y}{5}\) = \(\dfrac{z}{7}\) và 3x - 2z =15b)\(\dfrac{x}{5}\) = \(\dfrac{4}{3}\) = \(\dfrac{z}{2}\) và 2x -3y =100c)\(\dfrac{x}{-3}\) = \(\dfrac{4}{-5}\) \(\dfrac{z}{-4}\) và 3z -2x =36d) \(\dfrac{x}{2}\) = y = \(^{\dfrac{z}{3}}\) và 3x -2 + 4z...
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a )\(\dfrac{x}{3}\) = \(\dfrac{y}{5}\) = \(\dfrac{z}{7}\) và 3x - 2z =15
b)\(\dfrac{x}{5}\) = \(\dfrac{4}{3}\) = \(\dfrac{z}{2}\) và 2x -3y =100
c)\(\dfrac{x}{-3}\) = \(\dfrac{4}{-5}\) \(\dfrac{z}{-4}\) và 3z -2x =36
d) \(\dfrac{x}{2}\) = y = \(^{\dfrac{z}{3}}\) và 3x -2 + 4z =16
a,Áp sụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\\\Rightarrow x=-3.3=-9\\ \Rightarrow y=-3.5=-15\\ \Rightarrow z=-3.7=-21 \)
a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x}{9}=\dfrac{2z}{14}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\) (Vì 3x-2z=15)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-3\\\dfrac{y}{5}=-3\\\dfrac{z}{7}=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-21\end{matrix}\right.\)
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b) Ta có: \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x}{10}=\dfrac{3y}{9}=\dfrac{2x-3y}{10-9}=\dfrac{100}{1}=100\) (Vì 2x-3y=100)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=100\\\dfrac{y}{3}=100\\\dfrac{z}{2}=100\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=500\\y=300\\z=200\end{matrix}\right.\)
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c) Ta có: \(\dfrac{x}{-3}=\dfrac{y}{-5}=\dfrac{z}{-4}=\dfrac{3z}{-12}=\dfrac{2x}{-6}=\dfrac{3z-2x}{\left(-12\right)-\left(-6\right)}=\dfrac{36}{-18}=-2\) (Vì 3z-2x=36)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-3}=-2\\\dfrac{y}{-5}=-2\\\dfrac{z}{-4}=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\\z=8\end{matrix}\right.\)
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