![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(15x=10y\Leftrightarrow x=\frac{10y}{15}\left(1\right)\)
\(10y=6z\Leftrightarrow z=\frac{10y}{6}\left(2\right)\)
Thay (1) vào (2) vào biểu thức \(x.y.z=-1920\);ta được :
\(\frac{10y}{15}\cdot y\cdot\frac{10y}{6}=-1920\)
\(\Leftrightarrow\frac{100y^3}{90}=-1920\Leftrightarrow100y^3=-172800\Leftrightarrow y^3=-1728\Leftrightarrow y=-12\)
Với \(y=-12\Rightarrow x=\frac{10.\left(-12\right)}{15}=-8;z=\frac{10.\left(-12\right)}{6}=-20\)
Vậy \(x;y;z=\left\{-8;-12;-20\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(15x=10y=6z\)
\(\Leftrightarrow\)\(\frac{15x}{30}=\frac{10y}{30}=\frac{6z}{30}\)
\(\Leftrightarrow\)\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)\(\Rightarrow\)\(\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Suy ra \(xyz=-1920\)\(\Leftrightarrow\)\(2k.3k.5k=-1920\)
\(\Leftrightarrow\)\(30k=-1920\)
\(\Leftrightarrow\)\(k=\frac{-1920}{30}\)
\(\Leftrightarrow\)\(k=-64\)
Do đó :
\(x=2k=2.\left(-64\right)=-128\)
\(y=3k=3.\left(-64\right)=-192\)
\(z=5k=5.\left(-64\right)=-320\)
Vậy \(x=-128\)\(;\)\(y=-192\) và \(z=-320\)
Chúc bạn học tốt ~
Cảm ơn bạn nhiều nha !
Chúc bạn học tốt !
Bạn kết bạn với mình nhé !
![](https://rs.olm.vn/images/avt/0.png?1311)
15x = -10y = 6z
<=> \(\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\)
<=> \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{5}\)
Đặt \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=-3k\\z=5k\end{cases}}\)
Ta có: xyz = -30000
=> 2k.(-3k).5k = -30000
=> -30k3 = -30000
=> k3 = 1000
=> k = 10
=> x = 20, y = -30, z = 50
Vì 15x = -10y = 6z => \(\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\) => \(\frac{x}{2}=\frac{-y}{3}=\frac{z}{5}\)
Đặt : \(\frac{x}{2}=\frac{-y}{3}=\frac{z}{5}=k\), ta có : x = 2k ; y = (-3).k ; x = 5k
=> x.y.z = 2 .k. ( -3 ). k.5.k = -30.k3 = -30000
=> k3 = 1000 => k = 10 => x = 10. 2 = 20
=> y = 10. ( - 3 ) = -30
=> z = 10.5 = 50
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 15x = 6z
=> x = 6/15z
-10y = 6z
=> y= -3/5z
=> xyz = -30000
<=> (6/15z) . (-3/5z) . z = -30000
<=> z^3 .( -6/25) = -30000
<=> z^3 = 125000
<=> z = 50
=> y = -30
=> x = 20
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
15x = -10y
=> 3.x = -2.y => x/-2 = y/3 [1]
-10y = 6.z
=> -5.y = 3.z => y/3 = z/-5 [2]
Từ [1] và [2] => x/-2 = y/3 = z/-5
Đặt x/-2= y/3 = z/-5 = k
=> x= -2k ; y= 3k ; z= -5k
=> xyz = 30. k^3 = 30000 => k^3 = 1000 => k = 10
=> x= -20 ; y = 30 ; z= -50
Vậy x= -20 ; y= 30 ; z= -50
![](https://rs.olm.vn/images/avt/0.png?1311)
\(15x=-10y\) => \(\frac{x}{-10}=\frac{y}{15}\) => \(\frac{x}{-2}=\frac{y}{3}\)
\(-10y=6z\) => \(\frac{y}{6}=\frac{z}{-10}\) => \(\frac{y}{3}=\frac{z}{-5}\)
=> \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{5}\)
=> \(\left(\frac{x}{2}\right)^3=\left(\frac{y}{-3}\right)^3=\left(\frac{z}{5}\right)^3=\frac{xyz}{2.-3.5}=\frac{-30000}{-30}=1000\)
=> x = 20
y = -30
z = 50
Chúc bạn làm bài tốt
\(15x=-10y=6z\Rightarrow\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\)
\(\Rightarrow\)\(\frac{x}{2}=\frac{y}{-3}=\frac{z}{5}\)
Đặt \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{5}=n\)
\(\Rightarrow x=2n,y=-3n,z=5n\)
\(\Rightarrow xyz=2n.-3n.5n\)
\(=-30n^3=-30000\Rightarrow n^3=-1000=-10^3\)
\(\Rightarrow n=-10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\&xy=1200\)
\(\Leftrightarrow\dfrac{15}{20}=\dfrac{x-9}{y-12}\Leftrightarrow\dfrac{3}{4}=\dfrac{x-9}{y-12}\)
\(\Rightarrow\dfrac{9}{12}=\dfrac{x-9}{y-12}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{9}{12}=\dfrac{x-9}{y-12}=\dfrac{x-9+9}{y-12+12}=\dfrac{x}{y}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{xy}{y^2}=\dfrac{x^2}{xy}\)
Từ \(\dfrac{3}{4}=\dfrac{xy}{y^{^2}}\Rightarrow\dfrac{3}{4}=\dfrac{1200}{y^2}\Rightarrow y^2=1200.\dfrac{4}{3}=1600\)
\(\Rightarrow y=\sqrt{1600}=\pm40\)
+ TH1: \(y=40\Rightarrow x=30\)
\(\dfrac{15}{x-9}=\dfrac{40}{z-24}\Rightarrow z=80\) (tự giải pt)
+ TH2: \(y=-40\Rightarrow x=-30\)
\(\dfrac{15}{x-9}=\dfrac{40}{z-4}\Rightarrow z=-80\) (tự giải pt)
Vậy, các cặp \(\left(x;y;z\right)\) thỏa mãn là \(\left(30;40;80\right)\&\left(-30;-40;-80\right)\)
\(b.15x=-10y=6z\&xyz=30000\)
\(\Rightarrow\left\{{}\begin{matrix}15x=-10y\\-10y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-10}=\dfrac{y}{15}\\\dfrac{y}{6}=\dfrac{z}{-10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-20}=\dfrac{y}{30}\\\dfrac{y}{30}=\dfrac{z}{-50}\end{matrix}\right.\Rightarrow\dfrac{x}{-20}=\dfrac{y}{30}=\dfrac{z}{-50}\)
Đặt \(\dfrac{x}{-20}=\dfrac{y}{30}=\dfrac{z}{-50}=k\Rightarrow x=-20k;y=30k;z=-50k\)
\(\Rightarrow xyz=30000\Rightarrow-20k.30k.\left(-50k\right)=30000\Rightarrow30000k^3=30000\)
\(\Rightarrow k^3=1\Rightarrow k=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=30\\z=-50\end{matrix}\right.\)