1. illegal

2. traffic jam

3. seatbelt

4. safely

5. railway station

6. safety

7. traffic signs

8. helicopter

9. tricycle

10. boat

22. satisfied

23. considerate

24. generosity

25. proud

26. activist

27. generously

28. proud

29. unpolluted

30. preventing

31. Environmentalists

32. disappointing

33. complaint

34. conservation

35. disappointingly

36. environmentalists

37. unpolluted

38. Preventing

39. complaint

conservationists

considering

polluting

đề đâu ạ

13:

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}sin\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{2pi}{33}\right)\cdot cos\left(\dfrac{4pi}{33}\right)\cdot cos\left(\dfrac{8pi}{33}\right)\cdot cos\left(\dfrac{16pi}{33}\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{4}\cdot sin\dfrac{4}{33}pi\cdot cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{8}\cdot sin\dfrac{8}{33}pi\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{16}\cdot sin\dfrac{16}{33}pi\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{3}\right)}\cdot\dfrac{1}{32}\cdot sin\dfrac{32}{33}pi\)

=1/32

10:

\(=\dfrac{1}{2}\left[cos100+cos60\right]+\dfrac{1}{2}\cdot\left[cos100+cos20\right]\)

=cos100+1/2*cos20+1/4

6:

sin6*cos12*cos24*cos48

=1/cos6*cos6*sin6*cos12*cos24*cos48

=1/cos6*1/2*sin12*cos12*cos24*cos48
=1/cos6*1/4*sin24*cos24*cos48

=1/cos6*1/8*sin48*cos48

=1/cos6*1/16*sin96

=1/16

7C

8D

9A

10C

7-C.

8-C.

9-A.

10-C.

7-C        8-C           9-A        10-A

21 paid (loại 2)

22 will be (loại 1)

23 wouldn't have conquered (loại 3)

24 believed (loại 2)

25 would you choose (loại 2)

26 had scored (loại 3)

27 would certainlu accept ( loại 2)

28 had received (loại 3)

29 leave - calls (loại 0)

30 had (loại 2)

31 didn't drink (loại 2)

32 had turned on (loại 3)

33 had had (loại 3)

34 would have spoke (loại 3)

35 sends (loại 1)

36 wouldn't do (loại 2)

37 hadn't heard (loại 3)

3:

\(=\dfrac{2}{1+cotx-tanx-1}=\dfrac{2}{cotx-tanx}\)

\(=2:\left(\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}\right)=2:\dfrac{cos^2x-sin^2x}{sinx\cdot cosx}\)

\(=\dfrac{sin2x}{cos2x}\)

=tan2x

4:

\(=\left(1-\dfrac{1}{cot^2x}\right)\cdot cotx=cotx-\dfrac{1}{cotx}=\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}\)

\(=\dfrac{cos^2x-sin^2x}{sinx\cdot cosx}=\dfrac{cos2x}{\dfrac{1}{2}\cdot2\cdot sinx\cdot cosx}=\dfrac{cos2x}{sin2x}\cdot2\)

6:

\(=\dfrac{\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}}{cos2x}=\dfrac{cos^2x-sin^2x}{sinx\cdot cosx}:cos2x=\dfrac{1}{sinx\cdot cosx}\)

g) \(\left\{{}\begin{matrix}x-1\ge0\\2-\sqrt{x-1}\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\sqrt{x-1}\le2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\-4\le x-1\le4\end{matrix}\right.\)

\(\Leftrightarrow1\le x\le5\)

h) \(\left\{{}\begin{matrix}\dfrac{2x-4}{5-x}\ge0\\5-x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-4\ge0\\5-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-4\le0\\5-x< 0\end{matrix}\right.\end{matrix}\right.\\x\ne5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}5>x\ge2\left(tm\right)\\5< x\le2\left(vl\right)\end{matrix}\right.\\x\ne5\end{matrix}\right.\)

\(\Leftrightarrow5>x\ge2\)

i) \(x^2-8x-9\ge0\)\(\Leftrightarrow\left(x-4\right)^2-25\ge0\Leftrightarrow\left(x-4\right)^2\ge25\)

\(\Leftrightarrow-5\ge x-4\ge5\)\(\Leftrightarrow-1\ge x\ge9\)

j) \(2x-x^2>0\)

\(\Leftrightarrow\left(x-1\right)^2< 1\)

\(\Leftrightarrow-1< x-1< 1\Leftrightarrow0< x< 2\)

a: ĐKXĐ: \(2\le x\le4\)

b: ĐKXĐ: x>0

c: ĐKXĐ: \(x< \dfrac{1}{3}\)

g: ĐKXĐ: \(1\le x\le4\)

h: ĐKXĐ: \(2\le x< 5\)

i: ĐKXĐ: \(\left[{}\begin{matrix}x\ge9\\x\le-1\end{matrix}\right.\)

\(0< x< 2\)