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đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow x=2k,y=3k,z=5k\)
Ta có: \(xyz=-30\Rightarrow2k.3k.5k=-30\\ \Rightarrow30.k^3=-30\\ \Rightarrow k^3=-1\\ \Rightarrow k=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-2\\y=-3\\z=-5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(x\div y\div z=2\div3\div5\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\left(k\inℝ\right)\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Mà xyz = 30 \(\Rightarrow30k^3=30\Rightarrow k=1\)
\(\Rightarrow\hept{\begin{cases}x=2\\y=3\\z=5\end{cases}}\)
Theo bài ra ta có :
x : y : z = 2 : 3 : 5 và x.y.z = 30
=> \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
Gọi\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)=>\(\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Thay x=2k; y=3k; z=5k vào x.y.z = 30 ta có :
2k.3k.5k=30
<=> k3.30=30
=> k3=1
=> k=1
=>\(\hept{\begin{cases}x=2.1=2\\y=3.1=3\\z=5.1=5\end{cases}}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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a.
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)
Thế vào \(2x+y-z=81\)
\(\Rightarrow2.5k+3k-4k=81\)
\(\Rightarrow9k=81\)
\(\Rightarrow k=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)
b.
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)
Thế vào \(5x-y+3z=124\)
\(\Rightarrow5.3k-5k+3.2k=124\)
\(\Rightarrow16k=124\)
\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)
c.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Thế vào \(xyz=810\)
\(\Rightarrow2k.3k.5k=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải :
a) Đặt \(\frac{x}{5}=\frac{y}{4}=\frac{z}{2}=k\)
\(\Leftrightarrow\hept{\begin{cases}x=5k\\y=4k\\z=2k\end{cases}}\)
Ta có : \(xyz=40k^3=240\)
\(\Leftrightarrow k^3=6\)
\(\Leftrightarrow k=\sqrt[3]{6}\)
\(\Leftrightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{2}=\sqrt[3]{6}\)
\(\Leftrightarrow\hept{\begin{cases}x=5\sqrt[3]{6}\\y=4\sqrt[3]{6}\\z=2\sqrt[3]{6}\end{cases}}\)
Vậy....
b) \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{9}=\frac{y}{6}\)
Ta cũng có \(\frac{y}{3}=\frac{z}{2}\Leftrightarrow\frac{y}{6}=\frac{z}{4}\)
Khi đó : \(\frac{x}{9}=\frac{y}{6}=\frac{z}{4}=\frac{x-y+z}{9-6+4}=\frac{21}{7}=3\)
\(\Leftrightarrow\hept{\begin{cases}x=27\\y=18\\z=12\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=k\)
\(\Rightarrow xyz=5.2.\left(-3\right).k=-30k=240\Rightarrow k=-8\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-8\right).5=-40\\y=\left(-8\right).2=-16\\z=\left(-8\right).\left(-3\right)=24\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Đ}\text{ặt}:\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
\(\text{Có : }xyz=\left(-30\right)\)
\(2k.3k.5k=\left(-30\right)\)
\(30k^3=\left(-30\right)\)
\(k^3=\left(-1\right)\)
\(k=\left(-1\right)\)
\(\Rightarrow\hept{\begin{cases}x=2k=2\\y=3k=3\\z=5k=5\end{cases}}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Thay vào ta có :
\(2k.3k.5k=-30\)
\(30.k^3=-30\)
\(k^3=-1\)
\(k=-1\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=-1\\\frac{y}{3}=-1\\\frac{z}{5}=-1\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-2\\y=-3\\z=-5\end{cases}}\)