bài tập: tìm x, y, z, biết:
a)x : y : z = 3 : 5 : (-2) và 5x - y + 3z = -16
b) x + y + z = 5,2 và \(\dfrac{x}{2}\)= \(\dfrac{y}{-3}\); \(\dfrac{z}{3}\)=\(\dfrac{y}{4}\)
c) \(\dfrac{x+3}{5}\)=\(\dfrac{y-2}{3}\)=\(\dfrac{z-1}{7}\) và 3x + 5y - 7z =32
e)\(\dfrac{x}{4}\) = \(\dfrac{y}{5}\) và xy = 80
f) \(\dfrac{x}{4}\)= \(\dfrac{y}{3}\) và x\(^2\) - y\(^2\) = 63
e, Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\left(k\in Z\right)\)
\(\Leftrightarrow x=4k,y=5k\) (1)
Theo bài ra ta có: xy = 80
Từ (1) \(\Rightarrow4k.5k=80\Rightarrow20.k^2=80\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+ Với k = 2 \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
+ Với k = -2 \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)
Vậy \(\left(x,y\right)\in\left\{\left(8,10\right);\left(-8,-10\right)\right\}\)
a) \(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=\dfrac{-16}{4}=-4\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{z}{-2}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)
a, Ta có: \(x:y:z=3:5:\left(-2\right)\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{-z}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=-\dfrac{z}{2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5+\left(-6\right)}\)\(=\dfrac{-16}{4}=-4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{-z}{2}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-4.3=-12\\y=-4.5=-20\\-z=-4.2=-8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)
Vậy x = -12, y = -20, z = 8
