3 A

4 D

5 A

6 C

7 C

8 A

9 A

10 C

11 C

12 A

13 A

14 B

15 A

16 D

17 D

18 C

19 D

20 D

21 C

22 C

23 C

24 C

25 C

Ex2

1 remember 

2 read

3 be repaired

4 focus

5 not enter

6 apoligize

7 attend

8 sit

9 think

10 drive

11 be taken

12 reduce

13 allow

14 regulated

15 swim

16 monitor

17 out

18 be invited

19 send

20 work 

Bài 2: 

a: x=60

b: x=14

Bài 3: 

Ta có: \(\sqrt{4x^2-12x+9}=4\)

\(\Leftrightarrow\left|2x-3\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=7\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Bài 2: 

a: Ta có: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

\(=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)

\(=3-2\sqrt{2}+3+2\sqrt{2}\)

=6

b: Ta có: \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\)

\(=2\sqrt{2}\)

Bài 3:

a: Xét ΔBDC và ΔADH có

DB=DA

góc BDC=góc ADH

DC=DH

=>ΔBDC=ΔADH

Xét ΔCEB và ΔAEK có

EC=EA

góc CEB=góc AEK

EB=EK

=>ΔCEB=ΔAEK

b: ΔDAH=ΔDBC

=>góc DAH=góc DBC và AH=BC

mà hai góc so le trong

nên AH//BC

Xét tứ giác ABCK có

E là trung điểm chung của AC và BK

nên ABCK là hình bình hành

=>AK//BC và AK=BC

AK//BC

AH//BC

=>H,A,K thẳng hàng

mà AK=AH

nên A là trung điểm của KH

19d

20a

#\(Cụt cạp cạp\)

#\(yGLinh\)

1)

a) 4y²-4xy+x²= x²-4xy+4y²= (x-2y)²

b) 9x²-12xy+4y²= (3x)²-2.3x.2y+(2y)²= (3x-2y)²

c) 16x²-25=(4x)²-5²= (4x-5)(4x+5)

d) 1-9y²= 1²-(3y)²=(1-3y)(1+3y)

g) x³-27y³= (x-3y)(x²+3xy+9y²)

h) 64 + 8x³=(4+2x)(16+8x+4x²)

i) x⁴+4y⁴-4x²y²= (x²)²-2x²2y²+(y²)²=(x²-y²)²

Part 1

1.girls

2.are

3.reading

4.boy

5.she's

Part 2

2. No, they aren't. Some girls are making puppets

3. They are red, blue and yellow

4. The girl is reading a text

5. The teacher is in the office

6. Yes, she is reading a book

6.No, they aren't

7.They are painting red, blue and yellow.

8.The girl is reading a text.

9.Our teacher is in the office.

10.Yes, she is.