Ta có : 

\(3x=2y\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{2x}{4}\)

ADTCDTSBN , ta có : 

\(\frac{x}{2}=\frac{y}{3}=\frac{2x}{4}=\frac{y-2x}{3-4}=\frac{5}{-1}=-5\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=-5\\\frac{y}{3}=-5\end{cases}\Rightarrow\hept{\begin{cases}x=-5.2=-10\\y=-5.3=-15\end{cases}}}\)

Vậy \(x=-10;y=-15\)

\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(x^3-3^3+x\left(2^2-x^2\right)=1\)

\(x^3-27+4x-x^3=1\)

\(4x-27=1\)

\(4x=28\)

\(x=7\)

Vậy x = 7

\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(\Rightarrow x^3-3^3+x\left(2^2-x^2\right)=1\)

\(\Rightarrow x^3-27+4x-x^3=1\)

\(\Rightarrow4x-27=1\)

\(\Rightarrow4x=28\)

\(\Rightarrow x=7\)

Vậy \(x=7\)

Ta có \(xy\left(2x^2+1\right)-2x\left(2y^2+1\right)+1=x^3y^3\)

<=>\(x\left(x^2y^3-2x^2y-y+4y^2+2\right)=1\)

=> \(x^2y^3-2x^2y-y+4y^2+2=\frac{1}{x}\)

Do VT là số nguyên với x,y nguyên

=> \(\frac{1}{x}\)nguyên => \(x=\pm1\)

+ \(x=1\)=> \(y^3-3y+4y^2+1=0\)( không có nghiệm nguyên)

+ x=-1

=> \(y^3-3y+4y^2+3=0\)( không có nghiệm nguyên )

=> PT vô nghiệm 

Vậy PT vô nghiệm 

xy+3x-7y=21

x(y+3)-7y+21=0

x(y+3)-7(y+3)=0

(y+3).(x-7)=0(còn lại tự làm ha)

Ex 1:

1. listen

2. rains

3. will recognise

4. don’t want

5. does/ begin

6. will not

7. does/ watches

8. plant

9. play

10. leaves

Ex 2:

1. watching

2. listening

3. to buy

4. to speak

5. making

6. to eat

7. working

8. to call

9. to build

10. doing