12x=-15y=10z và xyz=120
12x = 15y = 28z và xyz = 14700
\(12x=15y=28z\)
\(\Rightarrow\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{15}\)
\(\Rightarrow\dfrac{x^3}{35^3}=\dfrac{y^3}{28^3}=\dfrac{z^3}{15^3}=\dfrac{xyz}{35.28.15}=\dfrac{14700}{14700}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=35\\y=28\\z=15\end{matrix}\right.\)
\(12x=15y=28z\Rightarrow\dfrac{x}{\dfrac{1}{12}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{28}}\)
Đặt \(\dfrac{x}{\dfrac{1}{12}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{28}}=k\Rightarrow x=\dfrac{1}{12}k;y=\dfrac{1}{15}k;z=\dfrac{1}{28}k\)
\(xyz=14700\\ \Rightarrow\dfrac{1}{12}k\cdot\dfrac{1}{15}k\cdot\dfrac{1}{28}k=14700\\ \Rightarrow\dfrac{1}{5040}k^3=14700\\ \Rightarrow k^3=74088000\\ \Rightarrow k=420\\ \Rightarrow\left\{{}\begin{matrix}x=35\\y=28\\z=15\end{matrix}\right.\)
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và -x+y+z= -120
\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}\) và x+y+z=48
\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) =>
Tìm x , y ,z biết :
a) \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\) và \(-x+y+z=-120\)
b) \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}\) và \(xyz=20\)
c) \(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}\) và \(x+y+z=48\)
b) Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)
\(\Rightarrow x=12k,y=9k,z=5k\)
\(xyz=20\)
\(\Rightarrow12k.9k.5k=20\)
\(\Rightarrow540k^3=20\)
\(\Rightarrow k^3=\frac{1}{27}\)
\(\Rightarrow k=\frac{1}{3}\)
Khi \(k=\frac{1}{3}\)
\(\Rightarrow\frac{x}{12}=\frac{1}{3}\Rightarrow x=4\)
\(\frac{y}{9}=\frac{1}{3}\Rightarrow y=3\)
\(\frac{z}{5}=\frac{1}{3}\Rightarrow z=\frac{5}{3}\)
Vậy x = ..... ; y = ............ ; z = .............
12x- 15y/7= 20z-12x/9=15y-20z/11 và x+y+z= 48
\(\frac{12x-5y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20x}{7+9+11}=0\)\(=0\)
=> 12x - 15y =0
=> 12x = 15y
=> \(\frac{x}{15}=\frac{y}{12}\)
=> \(\frac{x}{60}=\frac{y}{48}\)
20z - 12x = 0
=> 20z = 12x
=> \(\frac{x}{20}=\frac{z}{12}\)
=> \(\frac{x}{60}=\frac{z}{36}\)
=> \(\frac{x}{60}=\frac{y}{48}=\frac{z}{36}=\frac{x+y+z}{60+48+36}=\frac{48}{144}=\frac{1}{3}\)
=> x = 1 . 60 : 3 = 20
y = 1 . 48 : 3 = 16
z = 1 . 36 : 3 = 12
12x-15y/7=20z-12x/9=15y-20x/11 và x+y+z=48
tìm x,y,z biết: 12x-15y/7= 20z-12x/9= 15y-20z/11 và x+y+z=48
tìm nghiệm nguyên
6x + 15y + 10z = 3
Tìm nghiệm nguyên của PT: 6x+15y+10z=3
tìm x, ý, z biết 12x-15y/7=20z-12x/9=15y-20z/11 và x+y+z= 48