1. Phong and Hoang / go / school / together / Hoang 's bicycle / every morning //

=> Phong and Hoang go to school together by Hoang's bicycle every morning.

2. I / would like / borrow / books / please //

- When was The Lonely Giraffe made?

- I don't know. But I first watched it on TV last year.

1. Road USER should obey the traffic rules strictly USE

2. We should be CAREFUL drivers to avoid having accident. CARE

3. The student are LEARNING about road signs. LEARN

4. They are PROHIBITIVE signs PROHIBIT

1.I get to work in half an hour.
It takes me half an hour to get to work.

2. Do you have a cheaper computer than this?
Is this the cheapest computer?
3. How much is this dictionary?
How much does this dictionary cost?

4. It isn’t important for you to finish the work today.
You don’t need to finish the work today.
5. There are over eight hundred stamps in Tim’s collection.
Tim’s collection has over eight hundred stamps.

1. There used to... a building here but it was knocked down.

A. be B. is C.being D.was

2. Traffic accident can be...if people obey the rules.

a. prevented b. taken c. left d. made

3. You should look right and left when you go...the road.

a. along b. up c. across d. down

4. Tim used to... puzzle in his freetime.

a. dose b. did c. do d. doing

Gọi \(M\left(x;y\right)\) thì theo tính chất HBH ta có AB = CD, AD = BC

Áp dụng điều trên ta sẽ lập hệ như sau :

\(\left\{\begin{matrix}\sqrt{\left(1+2\right)^2+3^2}=\sqrt{\left(x-3\right)^2+y^2}\\\sqrt{\left(x-1\right)^2+\left(y-3\right)^2}=\sqrt{\left(3+2\right)^2}\end{matrix}\right.\)

Giải hệ trên tìm ra x,y là tọa độ của M nhé :)

Độ dài của hai điểm trên mặt phẳng tọa độ tính bằng công thức \(AB=\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2}\)

2.b

\(\left|x\right|.\sqrt{1-x^2}=\sqrt{x^2\left(1-x^2\right)}\le\frac{x^2+1-x^2}{2}\)

\(\Rightarrow\left|x\right|\sqrt{1-x^2}\le\frac{1}{2}\)

hay \(-\frac{1}{2}\le x\sqrt{1-x^2}\le\frac{1}{2}\)

Bạn tự tìm được rồi nhé :)

2. a) Áp dụng BĐT Bunhiacopxki, ta có

\(\left(\frac{1}{x}+\frac{4}{y}\right)\left(x+y\right)\ge\left[\left(\sqrt{\frac{1}{x}.x}\right)^2+\left(\sqrt{\frac{4}{y}.y}\right)^2\right]=\left(1^2+2^2\right)\)

\(\Rightarrow\frac{1}{x}+\frac{4}{y}\ge1\)

Đẳng thức xảy ra khi \(\left\{\begin{matrix}\frac{1}{x^2}=\frac{4}{y^2}\\x+y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{\begin{matrix}x=\frac{10}{3}\\y=\frac{5}{3}\end{matrix}\right.\)

 Có thể nói rõ hơn ở đề bài là M thuộc canh AB mà AM x 3 = AB. Nếu chỉ cho AM x 3 = AB thì có rất nhiều điểm M thỏa mãn

+) Nối M với C chia hình bình hành thành 2 phần có diện tích bằng nhau

S(AMC) = S(AMNC) /2 = 20 cm²

+) Tam giác CAB và CAM có chung chiều cao hạ từ C xuống AB; đáy AM = 1/3 đáy AB 

=> S(CAB) = 3 x S(AMC) = 30 cm²

+) S(ABCD) = 60 x 2 = 120 cm²

=> BD = S(ABCD) x 2 : AC = 120 x 2 : 24 = 10 cm