Question

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a; \(\frac{x}{4}=\frac{y}{7}=\frac{z}{5}\)và yz - xy - z²= -72

b; \(\frac{x}{2}=\frac{y}{7}=\frac{z}{8}\) và 2x² + xy - xz = 54

c; \(\frac{x+3}{5}=\frac{y-4}{3}=\frac{z-5}{2}\) và 2x - 3y - z = -26

Answer 1

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a, Đặt \(\frac{x}{4}=\frac{y}{7}=\frac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=7k\\z=5k\end{matrix}\right.\)

Mà \(yz-xy-z^2=-72\)

\(\Rightarrow35k^2-28k^2-25k^2=-72\\ \Rightarrow k^2\left(35-28-25\right)=-72\\ k^2\cdot\left(-18\right)=-72\\ \Rightarrow k^2=4\\ \Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

Với k = 2

\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot2=8\\y=7\cdot2=14\\z=5\cdot2=10\end{matrix}\right.\)

Với k = -2

\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot\left(-2\right)=-8\\y=7\cdot\left(-2\right)=-14\\z=5\cdot\left(-2\right)=-10\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)\in\left\{\left(8;14;10\right);\left(-8;-14;-10\right)\right\}\)

b, Đặt \(\frac{x}{2}=\frac{y}{7}=\frac{z}{8}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=7k\\z=8k\end{matrix}\right.\)

Mà \(2x^2+xy-xz=54\)

\(\Rightarrow8k^2+14k^2-16k^2=54\\ \Rightarrow k^2\left(8+14-16\right)=54\\ \Rightarrow k^2\cdot6=54\\ \Rightarrow k^2=9\\ \Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)

Với k = 3

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=7\cdot3=21\\z=8\cdot3=24\end{matrix}\right.\)

Với k = -3

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot\left(-3\right)=-6\\y=7\cdot\left(-3\right)=-21\\z=8\cdot\left(-3\right)=-24\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)\in\left\{\left(6;21;24\right);\left(-6;-21;-24\right)\right\}\)

c, Đặt \(\frac{x+3}{5}=\frac{y-4}{3}=\frac{z-5}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k-3\\y=3k+4\\z=2k+5\end{matrix}\right.\)

Mà \(2x-3y-z=-26\)

\(\Rightarrow2\left(5k-3\right)-3\left(3k+4\right)-\left(2k+5\right)=-26\\ \Rightarrow10k-6-9k-12-2k-5=-26\\ \Rightarrow-k=-3\\ \Rightarrow k=3\\ \Rightarrow\left\{{}\begin{matrix}x=5\cdot3-3=12\\y=3\cdot3+4=13\\z=2\cdot3+5=11\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(12;13;11\right)\)