11. is buying

12. is not studying

13. Is she running

14. is eating

15. are you waiting

16. are not trying

17. are having

18. are travelling

19. is drinking

20. is speaking

Ex2

13. b

14. a

15. c

16. c

17. b/d

18. a

19. a

20. d

21. d

22. a

16. 

1 for

2 been

3 than

4 so

5 writes

6 more

7 at

8 take

9 the

10 of

11 but

12 not

chữ đẹp thế

3. Tin's grandma told him not to go out then because it was snowing severely.
4. My brother said to me water boiled at 100 degrees Celsius.
5. Louis said watching cartoons had been his number one hobby years before, but he were into horor movies then.
6. The librarian told Arinna the book she was looking couldn't be there.
7. Mom said her friend had invited her to a wedding ceremony so she would have to buy a new dress.
8. Jennie said she had to finish that assignment that day and send it to the teacher via e-mail the next day.
9. The girl said those dry flowers looked even more gorgeous than the ones she had received.
10. Mr. Thomas always says his children are well-behaved.

ko nhìn thấy j đâu e

pls help me ;-;

\(a,3\left(x-1\right)-4=2\left(x+1\right)-7\\ \Leftrightarrow3x-3-4=2x+2-7\\ \Leftrightarrow3x-7=2x-5\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ b,\left(x+1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

c, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{3}{x^2-9}\\ \Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{\left(x^2+6x+9\right)-\left(x^2-6x+9\right)-3}{\left(x-3\right)\left(x+3\right)}=0\\ \Rightarrow x^2+6x+9-x^2+6x-9-3=0\\ \Leftrightarrow12x-3=0\\ \Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

d, \(d,\dfrac{x-3}{4}-\dfrac{2x+3}{3}=\dfrac{5x-1}{3}+\dfrac{2x+9}{12}\\ \Leftrightarrow d,\dfrac{3\left(x-3\right)}{12}-\dfrac{4\left(2x+3\right)}{12}=\dfrac{4\left(5x-1\right)}{12}+\dfrac{2x+9}{12}\\ \Leftrightarrow3x-9-8x-12=20x-4+2x+9\\ \Leftrightarrow-5x-21=22x+5\\ \Leftrightarrow27x+26=0\\ \Leftrightarrow x=-\dfrac{26}{27}\left(tm\right)\)

a: \(2+\dfrac{3\left(x+5\right)}{8}>\dfrac{x-1}{4}\)

=>\(2+\dfrac{3}{8}x+\dfrac{15}{8}>\dfrac{1}{4}x-\dfrac{1}{4}\)

=>\(\dfrac{3}{8}x+\dfrac{31}{8}>\dfrac{1}{4}x-\dfrac{1}{4}\)

=>\(\dfrac{1}{8}x>-\dfrac{1}{4}-\dfrac{31}{8}=\dfrac{-33}{8}\)

=>x>-33

b: \(2x-x\left(2x+1\right)< =15-2x\left(x+2\right)\)

=>\(2x-2x^2-x< =15-2x^2-4x\)

=>x<=15-4x

=>5x<=15

=>x<=3

a: \(\Leftrightarrow2x^2-2x+3x+6-2x^2+3=0\)

=>x+9=0

hay x=-9

b: \(\Leftrightarrow10x-30-4x=10x-5\)

=>-4x=-5+30=25

=>x=-25/4

j

:)?

Có rồi nha