1: Xét ΔABC vuông tại A có 

\(\widehat{B}+\widehat{C}=90^0\)

hay \(\widehat{C}=30^0\)

Xét ΔABC vuông tại A có 

\(BC=\dfrac{AC}{\sin60^0}\)

\(=\dfrac{32\sqrt{3}}{3}\left(cm\right)\)

hay \(AB=\dfrac{16\sqrt{3}}{3}\left(cm\right)\)

V.

1. B

2. B

3. A

4. D

5. A

6. C

7. D

8. D

9. A

10. C

11. B

12. C

13. A

14. A

15. B

16. D

17. C

18. A

19. B

20. B

III,

1. C

2. A

3. C

4. D

5. D

6. B

7. D

8. B

9. D

10. A

11. B

Cậu nên tách ra và đăng từng bài nha

III

1 Did you attend the Ban Flower Festival in Dien Bien last year?

2 Is Mua Sap a popular folk dance of the Thai people?

IV

1 They live in Wewbley, in north London

2 Her name is Tracy

3 No, there isn't

4 There are 4 people in his family

V

1  - B

2 - C 

3 - d

4 - f

5 - a

6 - e

7 - g

a) \(D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)\(=\left(\dfrac{2}{x+2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)

\(=\left(\dfrac{2\left(x+2\right)}{\left(x+2\right)^2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=\dfrac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x+4-4}{\left(x+2\right)^2}:\dfrac{-x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x}{\left(x+2\right)^2}.\dfrac{\left(x-2\right)\left(x+2\right)}{-x}\)

\(=\dfrac{-2.\left(x-2\right)}{x+2}\)

\(x^2-5x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(P=\dfrac{-2.\left(x-2\right)}{x+2}\)

Thay \(x=2\), ta có:

\(P=\dfrac{-2.\left(2-2\right)}{2+2}\)

    \(=0\)

Thay \(x=3\), ta có:

\(P=\dfrac{-2.\left(3-2\right)}{3+2}\)

    \(=-\dfrac{2}{5}\)

D nguyên âm \(\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\left(x-2\right)< 0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}-2\left(x-2\right)>0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -2\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)

a:Ta có: \(D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)

\(=\dfrac{2x+4-4}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{-x}\)

\(=\dfrac{-\left(x-2\right)}{x+2}\)

b: Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

hay x=3

Thay x=3 vào D, ta được:

\(D=\dfrac{-\left(3-2\right)}{3+2}=-\dfrac{1}{5}\)

like playing soccer, don't you
goes to school late, doesn't he
can swim very well, can't you
is going to the party, isn't she
was published in Germany in 1550, wasn't it
are sold all over the world, aren't they
have been built this year, haven't they
was given a book, wasn't he 
were bought by Mrs Brown yesterday, weren't they
is used everyday, isn't it

Bài 1:

\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=3\end{matrix}\right.\)