Part 1

1 on

2 as

3 who

4 so

5 well

6 didn't

Part 2

1 - G

2 - A

3 - F

4 - B

5 - C

6 - D

Part 3 

1 have walked

2 learnt

3 watches

4 playing

5 be done

6 impressed

7 surprisingly

8 cosumption

Part4

1 reduce => reduces

2 whom => who

part 1

1 ON

2WHOM

3 WHO

4 SO

5 WELL

6 DIDM'T

part 3

1 walk

2 learnt

3 playing

4 impressive

5 suprising

Q = \(\dfrac{3\sqrt{x}}{x+1}\) (x \(\ge\) 0; x \(\ne\) 4)

Áp dụng BĐT Cô-si cho 2 số không âm x và 1 ta được:

\(\dfrac{x+1}{2}\ge\sqrt{x}\) (1)

\(\Leftrightarrow\) \(\dfrac{3\cdot\dfrac{x+1}{2}}{x+1}\ge\dfrac{3\sqrt{x}}{x+1}\) (x + 1 > 0 với mọi x \(\ge\) 0)

\(\Leftrightarrow\) \(\dfrac{6}{2\left(x+1\right)}\ge\dfrac{3\sqrt{x}}{x+1}\)

\(\Leftrightarrow\) \(\dfrac{3}{x+1}\ge\dfrac{3\sqrt{x}}{x+1}\) (*)

Dấu "=" xảy ra \(\Leftrightarrow\) x = 1 (TM)

Khi đó: \(\dfrac{3\sqrt{x}}{x+1}\le\dfrac{3}{1+1}=\dfrac{3}{2}\)

Vậy Q_Max = \(\dfrac{3}{2}\) khi và chỉ khi x = 1

Chúc bn học tốt!

3 raise money for sth: gây quỹ cho

4 be dependent on st/sb: phụ thuộc vào cái gì/ai

5 be treated for + bệnh: được điều trị bệnh gì

6 not believe a word of st: không tin 1 chút nào vào cái gì

7 the habit of + Ving: thói quen gì

8 be intent on sth: quyết tâm, dốc lòng phấn đấu

9 stand a chance of + Ving: có cơ hội làm gì

10 take (no) notice of sth: (không) chú ý đến cái gì

11 it's high time sb did st: đến lúc phải làm gì

\(tan\left(\dfrac{x}{2}\right)=\sqrt{3}\)

\(\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{2\pi}{3}+k2\pi\) (\(k\in Z\))

Part 1

1 if

2 after

3 that

4 so

5 should

6 from

Part 2

1b 2e 3a 4f 5c 6d

Part 3

1 conservationists

2 to take

3 hasn't provided

4 works

5 disappointed

6 achievement

7 polluted

8 dissolve

Part 4

1 will get -> get

2 amazing -> amazed

Part 5 

1 use fresh water economically

2 school because the bus

3 they should keep

4 which is in this

Part 6

1 fond

2 watching

3 listened

4 or

Part 7 

1T 2F 3T 4F

\(2.a.Magie+Axitclohidric\rightarrow MagieClorua+Hidro\\ b.Mg+2HCl\rightarrow MgCl_2+H_2\\ c.m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ d.m_{HCl}=m_{MgCl_2}+m_{H_2}-m_{Mg}=47,5+1-12=36,5\left(g\right)\)

\(1.a.2Mg+O_2-^{t^o}\rightarrow2MgO\\ b.Fe+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2Ag\\ c.C_2H_4+3O_2-^{t^o}\rightarrow2CO_2+2H_2O\\ d.CuO+2HCl\rightarrow CuCl_2+H_2O\\ e.2Na+2H_2O\rightarrow2NaOH+H_2\\ f.4Al+3O_2-^{t^o}\rightarrow2Al_2O_3\)