\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\left(\dfrac{x}{2}\right)^3=\dfrac{xyz}{2.3.5}\Leftrightarrow\dfrac{x^3}{8}=\dfrac{240}{30}=8\Leftrightarrow x^3=64\Leftrightarrow x=4\Rightarrow\left\{{}\begin{matrix}y=6\\z=10\end{matrix}\right.\)

Đặt\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\)=k
=>x = 2k; y = 3k; z =5k
Mà x.y.z=240 => 2k.3k.5k=240
=>(k.k.k).(2.3.5)=240
=> \(k^3\) . 30 =240
=> \(k^3\) =240: 30
=> \(k^3\) = 8
=> k = \(\pm\) 2
Từ k=2 => x=2.2=4
k=-2=> x=-2.2=-4
Từ k=2 => y=2.3
k=-2=> y=-2.3=-6
Từ k=2=> z=2.5=10
k=-2=> z=-2.5=-10
Vậy x\(\in\pm\) 4
y\(\in\pm\) 6
z\(\in\pm\) 10

tìm x,y,z biết: x/2= y/3= z/5 với x.y.z=240

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

=> x = 2k, y = 3k, z = 5k

Ta có : x . y . z = 2k . 3k . 5k = 240

=> x . y . z = \(30k^3\)= 240

=> \(k^3=230:40=8\)

=> k = 2

=> x = 2k = 2 . 2 =4

=> y = 3k = 3 . 2 = 6

=> z = 5 . 2 = 10

Vậy x = 4, y = 6, z = 10

\(Đặt\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k,\)\(\) ta có: \(x=2k;y=3k;z=5k\)

Vì \(x.y.z=-240\Rightarrow2k.3k.5k=-240\)

\(\Rightarrow30k^3=-240\Rightarrow k^3=-240:30=-8\)

\(\Rightarrow k^3=\left(-2\right)^3\Rightarrow k=-2\)

\(\)Ta có:

\(x=2k\Rightarrow x=-2.2=-4\)

\(y=3k\Rightarrow y=-2.3=-6\)

\(z=5k\Rightarrow z=-2.5=-10\)

Vậy \(x=-4;y=-6;z=-10\)

Từ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x.y.y}{2.3.5}=\dfrac{-240}{30}\) = \(-8\)

=> \(\dfrac{x}{2}=\left(-8\right);\dfrac{y}{3}=\left(-8\right);\dfrac{z}{5}=\left(-8\right)\)

Với : \(\dfrac{x}{2}=\left(-8\right)\Rightarrow x=-16\)

Với:\(\dfrac{y}{3}=\left(-8\right)\Rightarrow y=-24\)

Với:\(\dfrac{z}{5}=\left(-8\right)\Rightarrow z=-40\)

a) Ta đặt: \(\frac{x}{4}=\frac{y}{3}=\frac{z}{-2}=k\)

\(\Rightarrow x=4k;y=3k;z=-2k\)

\(\Rightarrow xyz=\left(4.3.-2\right).k^3\)

\(\Rightarrow xyz=\left(-24\right).k^3\)

\(\Rightarrow k^3=240:\left(-24\right)=-10\)

\(\Rightarrow\)(đề sai, không ra số tự nhiên)

nếu đề cho là tìm thui thì là thuộc Z đó bạn

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...

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=k\)

\(\Rightarrow xyz=5.2.\left(-3\right).k=-30k=240\Rightarrow k=-8\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-8\right).5=-40\\y=\left(-8\right).2=-16\\z=\left(-8\right).\left(-3\right)=24\end{matrix}\right.\)

a) Đặt\(\frac{x}{5}=\frac{y}{2}=\frac{z}{3}=k.\)

Ta có : x = 5k ;  y = 2k ; z = 3k và xyz = 240

=> 5k . 2k . 3k = 240

=> k³ . 30 = 240

=> k³ = 8

=> k = 2

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=2\Leftrightarrow x=10\\\frac{y}{2}=2\Leftrightarrow y=4\\\frac{z}{3}=2\Leftrightarrow x=6\end{cases}}\)  

Vậy : x = 10; y = 4; z = 6

b) \(\frac{x}{4}=\frac{y}{3}=\frac{z}{2}\Rightarrow\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có :

\(\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{4}=\frac{x^2-y^2-z^2}{16-9-4}=\frac{12}{3}=4\)

Suy ra :

\(\frac{x^2}{16}=4\Leftrightarrow x^2=64\Leftrightarrow x=\pm8\)

\(\frac{y^2}{9}=4\Leftrightarrow y^2=36\Leftrightarrow y=\pm6\)

\(\frac{z^2}{4}=4\Leftrightarrow z^2=16\Leftrightarrow z=\pm4\)

Vậy \(\hept{\begin{cases}x=8\\y=6\\z=4\end{cases}}\)hoặc \(\hept{\begin{cases}x=-8\\y=-6\\z=-4\end{cases}}\)

c) \(\frac{x}{4}=\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{x^2+y^2+z^2}{16+9+25}=\frac{200}{50}=4\) 

Suy ra :

\(\frac{x^2}{16}=4\Leftrightarrow x^2=64\Leftrightarrow x=\pm8\)

\(\frac{y^2}{9}=4\Leftrightarrow y^2=36\Leftrightarrow y=\pm6\)

\(\frac{z^2}{25}=4\Leftrightarrow z^2=100\Leftrightarrow z=\pm10\)

Vậy :\(\hept{\begin{cases}x=8\\y=6\\z=10\end{cases}}\)hoặc \(\hept{\begin{cases}x=-8\\y=-6\\z=-10\end{cases}}\)