Từ \(\dfrac{2}{3}x=\dfrac{1}{2}y=2z\)

\(\Rightarrow\dfrac{2}{3}x.6=\dfrac{1}{2}y.6=2z.6\)

\(\Rightarrow4x=3y=12z\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=z\)

\(\Rightarrow\dfrac{3x}{9}=\dfrac{2y}{8}=\dfrac{z}{1}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{3x}{9}=\dfrac{2y}{8}=\dfrac{z}{1}=\dfrac{3x+2y+z}{9+8+1}=\dfrac{1}{18}\)

\(\Rightarrow\dfrac{y}{4}=\dfrac{1}{18}\Rightarrow y=\dfrac{1.4}{18}=\dfrac{2}{9}\)

\(\dfrac{2}{3x}=\dfrac{1}{2y}=\dfrac{2}{z}\)

\(\Rightarrow\dfrac{3x}{2}=\dfrac{2y}{1}=\dfrac{z}{2}=\dfrac{3x+2y+z}{2+1+2}=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{2y}{1}=\dfrac{1}{5}\)

\(\Rightarrow2y=\dfrac{1}{5}\)

\(\Rightarrow y=\dfrac{1}{10}\)

1) ADTCDTSBN, ta có:

 \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)= \(\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}\)= 4

* \(\frac{x}{3}=4\)=> x = 3 . 4 = 12

- \(\frac{y}{4}=4\)=> y = 4 . 4 = 16

* \(\frac{z}{5}=4\)=> z = 5 . 4 = 20

Vậy x = 12

       y = 16

       z = 20

x=12

y=16

z=20

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{2}{3x}=\frac{1}{2y}=\frac{2}{z}=\frac{2+1+2}{3x+2y+z}=\frac{5}{1}=5\)

\(\Rightarrow\frac{1}{2y}=5\Rightarrow2y=\frac{1}{5}\Rightarrow y=\frac{1}{5}:2=\frac{1}{10}\)

\(y=\frac{1}{10}\)                              

P = 5x² + 2y + 3x -z

thay x=1;y=-1;z=-2 ta có

P = 5 .1² + 2. (-1) + 3.1

P = 5 .1 + 2. (-1) + 3.1

P = 10

Tại \(x=1;y=-1;z=-2\) ta có:

\(P=5.1^2.\left(-1\right)+3.1-\left(-2\right)\)

\(P=5.1.\left(-1\right)+3.1-\left(-2\right)\)

\(P=-5+3+2=0\)

Vậy tại \(x=1;y=-1;z=-2\) vào biểu thức \(P\) là 0