a) \(2\overrightarrow{JA}+5\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Leftrightarrow2\overrightarrow{JA}+2\overrightarrow{JB}+3\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Leftrightarrow2\left(\overrightarrow{JA}+\overrightarrow{JB}\right)+3\left(\overrightarrow{JB}+\overrightarrow{JC}\right)=\overrightarrow{0}\)

\(\Leftrightarrow4\overrightarrow{JM}+6\overrightarrow{JN}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{JM}=-\dfrac{3}{2}\overrightarrow{JN}\)

\(\Rightarrow M;N:J\) thẳng hàng \(\Rightarrow\) Đúng

b) Sai (Chứng minh trên)

c) \(2\overrightarrow{IA}+3\overrightarrow{IC}=\overrightarrow{0}\)

\(\Leftrightarrow2\overrightarrow{IJ}+2\overrightarrow{JA}+3\overrightarrow{IJ}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Leftrightarrow5\overrightarrow{IJ}+2\overrightarrow{JA}+3\overrightarrow{JC}=\overrightarrow{0}\)

mà \(2\overrightarrow{JA}+5\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Rightarrow5\overrightarrow{IJ}-5\overrightarrow{JB}=\overrightarrow{0}\) (trừ vế với vế)

\(\Leftrightarrow\overrightarrow{IJ}=\overrightarrow{JB}\)

\(\Rightarrow J\) là trung điểm \(BI\Rightarrow\) Đúng

d) \(\overrightarrow{AE}=\dfrac{5}{7}\overrightarrow{AB}\)

\(\Leftrightarrow\overrightarrow{AJ}+\overrightarrow{JE}=\dfrac{5}{7}\overrightarrow{AJ}+\dfrac{5}{7}\overrightarrow{JB}\)

\(\Leftrightarrow\overrightarrow{JE}=-\dfrac{2}{7}\overrightarrow{AJ}+\dfrac{5}{7}\overrightarrow{JB}=\dfrac{2}{7}\overrightarrow{JA}+\dfrac{5}{7}\overrightarrow{JB}\left(1\right)\)

Ta lại có  \(2\overrightarrow{JA}+5\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{JC}=-\dfrac{2}{3}\overrightarrow{JA}-\dfrac{5}{3}\overrightarrow{JB}\left(2\right)\)

\(\left(1\right);\left(2\right)\) Ta thấy \(\dfrac{\dfrac{2}{7}}{-\dfrac{2}{3}}=\dfrac{\dfrac{5}{7}}{-\dfrac{5}{3}}=-\dfrac{3}{7}\)

\(\Rightarrow\overrightarrow{JE}=-\dfrac{3}{7}\overrightarrow{JC}\)

\(\Rightarrow C,E,J\) thẳng hàng \(\Rightarrow\) Đúng

\(\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}\)

\(\Rightarrow F^2=F_1^2+F_2^2\left(\widehat{AMB}=90^o\right)\)

\(\Rightarrow F=\sqrt{30^2+40^2}=500\left(N\right)\)

\(\Rightarrow\) Chọn D

\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3\left(mol\right)\)

Ta có: 56n_Fe + 127n_FeCl2 = 31 (1)

BTNT Fe: n_Fe + n_FeCl2 = n_FeCl3 = 0,3 (2)

Từ (1) và (2) ⇒ n_Fe = 0,1 (mol)

\(\Rightarrow\%m_{Fe}=\dfrac{0,1.56}{31}.100\%\approx18,06\%\)

Đáp án: A

 '' Sao Đến chân mới nhảy  =)) ''

24. The pupil who is sitting next to me is not clever.
25. The book that is on the top shelf is the one that I need.
26. The girl who is drinking coffee is my classmate.
27. The man who is in the room is angry.
28. The students who are having the final test look very nervous.
29. Bill Clinton, who is from Arkansas, is the forty-second president of the United States.
30. Sai Gon, which was founded in 1698, is one of the largest cities in Vietnam.
31. The picture that is on the wall was painted by my father.
32. My chair, which is made of wood, is expensive.
33. Nguyen Du, who was born in Ha Tinh province, is a great poet.
34. The student who is from China studies Vietnamese.
35. We are learning sentences that contain adjective clauses.
36. The girl whose eyes are bright is smart.
37. Saturday is the day when boys and girls go together.
38. This is the village where I was born.
39. I've sent him two letters, neither of which he has received.
40. Last year I visited a country that is very beautiful.

24 The pupil who is sitting next to me is not clever

25 The book that I need is on the top shelf

26 The girl who is drinking coffee is my classmate

27 The man who is in the room is angry

28 The students who are having final test look very nervous

29 Bill Clinton, who is from Arkansas, is the forty-second president of the United States

30 Sai Gon, which was founded in 1698, is the one of the largest cities in Vietnam

31 The picture which is on the wall was painted by my father

32 My chair, which is expensive, is made of wood

33 Nguyen Du, who was born in Ha Tinh province, is a great poet

34 The student who studies Vietnamese is from China

35 We are learning sentences that contain adjective clauses

1. The girl who chatted with you yesterday is my cousin.
2. The man who spent 13 minutes measuring our kitchen is very kind.
3. The architect who designed these flats has moved to HCM City.
4.  The young boy who rides an expensive motorbike is naughty.
5.  I'm reading the book which was written in 1996.
6.   Do you know the man who is sharing the flat with me?
7 . The young man who is talking to our teacher is Ba's brother.
8. We visited the monument which was built a hundred years ago.
9. The boys are interested in the game which is a popular game all over the world.
10. You have no need to tell me the reason why you are often late for work.
11. We visited the people who are homeless after the flood.
12. I don't like the films which have an unhappy ending.
13. I know the man whose son broke your windows.
14. The girl whose parents work with me is very kind.
15. She hasn't eaten anything, which makes her parents worried.
16. The student who sits next to me is from China.
17. I thanked the woman who had helped me.
18. Mr. Pike, who I am taking his course, is excellent.
19. Shakespeare wrote plays which people have enjoyed for four centuries.
20. This is the bank from which we borrowed the money.
21. Jack's car had broken down, so he had to take a bus.
22. Some boys who were arrested have been released.

3 câu đầu đúng r nhé

4 The young boy who rides an expensive motorbike is naughty

5 I'm reading the book which I bought in 1996

6 Do you know the man who is sharing the flat with me?

7 The young man who is talking to our teacher is Ba's brother

8 We visited the monument which was built a hundred years ago

C23 : A 

C41:  A 

C30: A 

Bài 2:

Gọi O là tâm của hình vuông ABCD

=>O là trung điểm chung của AC và BD

ABCD là hình vuông

=>\(AC^2=AB^2+BC^2=a^2+a^2=2a^2\)

=>\(AC=\sqrt{2a^2}=a\sqrt{2}\)

=>\(BD=AC=a\sqrt{2}\)

\(\left|\overrightarrow{AC}-\overrightarrow{BD}\right|\)

\(=\left|2\cdot\overrightarrow{AO}-2\cdot\overrightarrow{BO}\right|\)

\(=2\left|\overrightarrow{AO}+\overrightarrow{OB}\right|=2\cdot\left|\overrightarrow{AB}\right|=2\cdot AB=2a\)

\(\left|\overrightarrow{AB}-\overrightarrow{BC}-\overrightarrow{CD}-\overrightarrow{DA}\right|\)

\(=\left|\overrightarrow{AB}-\overrightarrow{AD}-\overrightarrow{BA}-\overrightarrow{DA}\right|\)

\(=\left|\overrightarrow{AB}+\overrightarrow{AB}+\overrightarrow{DA}-\overrightarrow{DA}\right|=\left|2\overrightarrow{AB}\right|=2\cdot AB=2a\)