a) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)và \(2x-y+3z=28\)

Theo tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x}{5}=\dfrac{2x-y+3z}{4-3+15}=\dfrac{28}{16}=1,75\)

\(\Rightarrow\dfrac{x}{2}=1,75\Rightarrow x=3,5\)

\(\dfrac{y}{3}=1,75\Rightarrow y=5,25\)

\(\dfrac{z}{5}=1,75\Rightarrow z=8,75\)

\(\dfrac{x}{7}\) = \(\dfrac{y}{6}\) = \(\dfrac{z}{3}\) 

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

\(\dfrac{x}{7}=\dfrac{y}{6}=\dfrac{z}{3}\) = \(\dfrac{x+y+z}{7+6+3}\) = \(\dfrac{-40}{16}\) = - \(\dfrac{5}{2}\)

\(x=-\dfrac{5}{2}.7\) = - \(\dfrac{35}{2}\); y = - \(\dfrac{5}{2}.6=-15\);  z = - \(\dfrac{5}{2}.3=-\dfrac{15}{2}\)

Vậy (\(x;y;z\)) = (- \(\dfrac{35}{2}\); -15;  - \(\dfrac{15}{2}\)) 

`#040911`

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

\(\dfrac{x}{7}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{x+y+z}{7+6+3}=\dfrac{-40}{16}=-\dfrac{5}{2}=-2,5\)

\(\Rightarrow\dfrac{x}{7}=\dfrac{y}{6}=\dfrac{z}{3}=-2,5\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2,5\cdot7=-17,5\\y=-2,5\cdot6=-15\\z=-2,5\cdot3=-7,5\end{matrix}\right.\)

Vậy, `x = -17,5`; `y = -15`; `z = -7,5.`

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...

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

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

Tiếc was bài này mk lm đc mà đang onl bằng Đt nên ko vào cx đc huhuhuh

từ \(\frac{x}{2}\)= \(\frac{y}{3}\)= \(\frac{z}{5}\) => \(\frac{x}{2}\)= \(\frac{y}{3}\) = \(\frac{z}{5}\)

= \(\frac{x.y.z}{2.3.5}\) = \(\frac{240}{30}\)= 8

=> x = 8 . 2       => x = 16

     y = 8 . 3             y = 24

     z = 8 . 5             z = 40

vậy x , y , z = 16 , 24 , 40