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

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

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

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

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

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

Lời giải:

Áp dụng TCDTSBN:

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

$\Rightarrow 3x=\frac{2}{5}; 2y=\frac{1}{5}; z=\frac{2}{5}$

$\Rightarrow x=\frac{2}{15}; y=\frac{1}{10}; z=\frac{2}{5}$

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:

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

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

1)x=7-2y nên x+2y=7

A=3x+6y+5=3x(x+2y)+5=3x7+5=26

2)y-x=y+(-x)=-x+y=-(x-y)=-10

Vậy x-y=10

Nguyễn Thị Ngọc Ánh: Mk chưa hiểu câu 1 cho lắm, bn giải kĩ hơn đc ko?