Tìm x,y,z biết x/12=y/9=z/5; xyz=20
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 9:
Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)
Vậy: (x,y,z,t)=(-10;6;34;-18)
Bài 11:
Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)
Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)
\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)
Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)
\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)
Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)
\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)
Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)
\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)
Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)
\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)
Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)
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
a, 5x = 8y => \(\frac{x}{8}=\frac{y}{5}\)
8y = 20z => 2y = 5z => \(\frac{y}{5}=\frac{z}{2}\)
=> \(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}=\frac{x-y-z}{8-5-2}=\frac{3}{1}=3\)
=> x = 24,y = 15,z = 6
b, \(\frac{6}{11}x=\frac{9}{2}y\)=> \(\frac{12x}{22}=\frac{99y}{22}\)=> 12x = 99y => 4x = 33y => \(\frac{x}{33}=\frac{y}{4}\)
\(\frac{9}{2}y=\frac{18}{5}z\)=> \(\frac{45y}{10}=\frac{36z}{10}\)=> 45y = 36z => 5y = 4z => \(\frac{y}{4}=\frac{z}{5}\)
=> \(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{120}{-24}=-5\)
=> x = -165 , y = -20 , z = -25
c, Đặt : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)=> x = 12k , y = 9k , z = 5k
=> xyz = 12k . 9k . 5k
=> xyz = 540k3
=> 540k3 =20
=> k3 = 20/540
=> k3 = 1/27
=> k = 1/3
Do đó : x= 4 , y = 3 , z = 5/3
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
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 :
\(12k.9k.5k=20\)
\(540.k^3=20\)
\(k^3=\frac{1}{27}\)
\(k^3=\left(\frac{1}{3}\right)^3\)
\(k=\frac{1}{3}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{12}=\frac{1}{3}\\\frac{y}{9}=\frac{1}{3}\\\frac{z}{5}=\frac{1}{3}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=\frac{5}{3}\end{cases}}\)
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