Bài 1: 
1. am
2. Does she play
3. go
4. Is she 
5. find
6. Are they 
7. rides 
8. is he
9. don't play
10. is not
11. are
12. Do we make 
13. is Harry study
14. Is it
15. are not 
16. don't like
17. are you
18. isn't
19. Is the dog eat?
20. is
Bài 2:
1. is running
2. buy
3. doesn't drink
4. are climbing
5. is crying
6. wear 
7. do you do 
8. eat 
9. sets off
10. are not going 

Bài 1: 
1. am
2. Does she play
3. go
4. Is she 
5. find
6. Are they 
7. rides 
8. is he
9. don't play
10. is not
11. are
12. Do we make 
13. is Harry study
14. Is it
15. are not 
16. don't like
17. are you
18. isn't
19. Is the dog eat?
20. is
Bài 2:
1. is running
2. buy
3. doesn't drink
4. are climbing
5. is crying
6. wear 
7. do you do 
8. eat 
9. sets off
10. are not going 

Nếu đúng bạn like giúp mk nhé

\(5,\\ 5\cdot625=25\cdot125\\ \Leftrightarrow\dfrac{5}{25}=\dfrac{125}{625};\dfrac{5}{125}=\dfrac{25}{625};\dfrac{625}{25}=\dfrac{125}{5};\dfrac{625}{125}=\dfrac{25}{5}\)

\(6,\)

\(a,\) Áp dụng t/c dtsbn: \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-21}{7}=-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\cdot2=-6\\y=-3\cdot5=-15\end{matrix}\right.\)

\(b,7x=3y\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{7}\)

Áp dụng t/c dtsbn: \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-4\cdot3=-12\\y=-4\cdot7=-28\end{matrix}\right.\)

\(7,\\ a:b:c:d=2:3:4:5\Leftrightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{d}{5}\)

Áp dụng t/c dtsbn: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{d}{5}=\dfrac{a+b+c+d}{2+3+4+5}=\dfrac{-42}{14}=-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-3\cdot2=-6\\b=-3\cdot3=-9\\c=-3\cdot4=-12\\d=-3\cdot5=-15\end{matrix}\right.\)

1. should ask

*must not ask là sai vì người quản lí mới nên hỏi vài câu hỏi trước khi thực hiện thay đổi

*have to ask (mang tính chủ quan) nên có làm hay không làm cũng được nên ta không chọn đáp án này

2. could prevent

Vì sau would, could là V_inf

\(5,\\ \left\{{}\begin{matrix}AB\perp AD\\CD\perp AD\end{matrix}\right.\Rightarrow AB//CD\\ \Rightarrow\widehat{ABC}+\widehat{BCD}=180^0\left(2.góc.TCP\right)\\ \Rightarrow60^0+\widehat{BCD}=180^0\\ \Rightarrow\widehat{BCD}=120^0\)

\(6,\\ a,\left\{{}\begin{matrix}a\perp AB\\a//b\end{matrix}\right.\Rightarrow b\perp AB\\ b,a//b\Rightarrow\widehat{C}+\widehat{D}=180^0\left(2.góc.TCP\right)\\ \Rightarrow\widehat{D}=180^0-120^0=60^0\\ c,\widehat{ICD}=\dfrac{1}{2}\widehat{ACD}=\dfrac{1}{2}\cdot120^0=60^0\left(t/c.phân.giác\right)\\ \Delta CID.có:\widehat{CID}+\widehat{IDC}+\widehat{ICD}=180^0\\ \Rightarrow60^0+60^0+\widehat{CID}=180^0\\ \Rightarrow\widehat{CID}=60^0\)