x/2=y/9=z/5 và xyz=20. tìm x; y; z
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)
\(\Rightarrow x=12k;y=9k;z=5k\)
Mà xyz = 20
\(\Rightarrow\)12k . 9k . 5k = 20
\(\Rightarrow\)540k3 = 20
\(\Rightarrow\)k3 = \(\frac{1}{27}\)
\(\Rightarrow\)k = ( -3 )
\(\Rightarrow\)x = -36 ; y = -27 ; z = -15
Ta có:
\(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\Leftrightarrow x=12k;y=9k;z=5k\) và \(xyz=20\)
\(\Rightarrow12k.9k.5k=20\)
\(\Rightarrow540k^3=20\Leftrightarrow k=\sqrt[3]{20:540}=\frac{1}{3}\)
\(\hept{\begin{cases}x=12.\frac{1}{3}=4\\y=9.\frac{1}{3}=3\\z=5.\frac{1}{3}=\frac{5}{3}\end{cases}}\)
Vậy x = 4; y = 3 ; z = 5/3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left[\begin{array}{nghiempt}xyz=20\\\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}xyz=20\\x=12k\\y=9k\\z=5k\end{array}\right.\)
\(\Rightarrow xyz=12k.9k.5k=540k^3\)
\(\Rightarrow20=540k^3\)
\(\Rightarrow k^3=\frac{20}{540}=\frac{1}{27}\Rightarrow k^3=\left(\frac{1}{3}\right)^3\Rightarrow k=\frac{1}{3}\)
\(\Rightarrow x=12k=12.\frac{1}{3}=4\)
\(\Rightarrow y=9k=9.\frac{1}{3}=3\)
\(\Rightarrow z=5k=\frac{5.1}{3}=\frac{5}{3}\)
TA CÓ X/12=Y/9=Z/5 =>X=12K;Y=9K;Z=5K
MÀ XYZ=20=>12K.9K.5K=20 HAY 540\(K^3\)=20
=>\(K^3\)=20/540=1/27=>\(K^3\)=\(\left(\frac{1}{3}\right)^3\)=>K=1/3
TỪ X/12=1/3=>X=4
Y/9=1/3=>Y=3
Z/5=1/3=>Z=5/3
VẬY X=4;Y=3;Z=5/3
TICK ĐÚNG CHO MIK NHA
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì x: y : z = 12 : 9 : 5 nên \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)
Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)
\(\Rightarrow\hept{\begin{cases}x=12k\\y=9k\\z=5k\end{cases}}\)
Thay vào ta có :
\(12k.9k.5k=20\)
\(540.k^3=20\)
\(k^3=\frac{1}{27}\)
\(k^3=\left(\frac{1}{3}\right)^3\)
\(k=\frac{1}{3}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{12}=\frac{1}{3}\\\frac{y}{9}=\frac{1}{3}\\\frac{z}{5}=\frac{1}{3}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=\frac{5}{3}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x:y:z=12:9:5 ->\(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)
Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\)= k -> x = 12*k ;y = 9*k ;z = 5*k (1)
thay (1) vào xyz=20 ta được : 12k + 9k
![](https://rs.olm.vn/images/avt/0.png?1311)
x/12=y/9=z/5 = k => x = 12k ; y = 9k ; z = 5k
Thay vào ta được:
12k.9k.5k = 20
540k3 = 20
k3 = 1/27
Vậy k = 1/3
x = 1/3 . 12 = 4
y = 9.1/3 = 3
z = 1/3 . 5 = 5/3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt x/12 = y/9 = z/5 = k ta có:
x = 12k
y = 9k
z = 5k
=> x.y.z = 12k.9k.5k
=> k^3.540=20
=> k^3 = 1/27
=> k^3= (1/3)^3
=> k = 1/3
x/12=1/3 => x=4
y/9= 1/3 => y=3
z/5=1/3 =. z=5/3
Gọi \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)
\(\Rightarrow x=12k;y=9k;z=5k\)
\(\Rightarrow xyz=12k.9k.5k=540k^3=20\)
\(\Rightarrow k^3=\frac{20}{540}=\frac{1}{27}=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow k=\frac{1}{3}\)
\(\Rightarrow\frac{x}{12}=\frac{1}{3}\Rightarrow x=\frac{1}{3}.12=4\)
\(\frac{y}{9}=\frac{1}{3}\Rightarrow y=\frac{1}{3}.9=3\)
\(\frac{z}{5}=\frac{1}{3}\Rightarrow z=\frac{1}{3}.5=\frac{5}{3}\)
Vậy \(x=4;y=3;z=\frac{5}{3}\)
uk lạ lắm thầy giáo giang r mà hỉu dc nhiu đó thâu
khó là chỗ k3 ó bạn