\(21_{10}=10101_2\)

1.A
2.A
3.B
4.C
5.B
6.C
7.A
8.A
9.B
10.A
11.B
12.A
13.C
14.B
15.B
16.A
17.A
18.A
19.A
20.C

g​iúp​ gì??

lỗi òiii

g: =>2x+1/2=0 hoặc 2x-3=0

=>x=-1/4 hoặc x=3/2

h: =>4x-5=0 hoặc 5/4x-2=0

=>x=5/4 hoặc x=2:5/4=2*4/5=8/5

i: =>(2x+1/2)^2=1/9

=>2x+1/2=1/3 hoặc 2x+1/2=-1/3

=>2x=-1/6 hoặc 2x=-5/6

=>x=-5/12; x=-1/12

k: =>(3x-1/2)^2=4/9

=>3x-1/2=2/3 hoặc 3x-1/2=-2/3

=>3x=7/6 hoặc 3x=-1/6

=>x=7/18; x=-1/18

i: =>7/6x=-10/3+1=-7/3

=>x=-2

m: =>x+1=4

=>x=3

5 he had done all his work, he went home

6 Steven bought a new motorbike, he had saved enough money

9 Mackenzie had written 4 best-sellers

10 were having dinner, there was a knock at the door

11 I was staying in Paris last summer

12 had told him off, I realized I was wrong

13 passed the exam, I had worked very hard

14 had considered what to study, I decided to major in Math

15 she was doing homework, her mom called her out to walk

16 bought a radio, he had checked the price

17 After they argued, they fought

18 she had a fatal accident, she had gone out for a walk

Bạn sai cta nhé, sắp not xắp

Tham khảo

1. A special kind of tea is sold here 

2. All the cars and trucks have been searched 

3. He was put in prison by the goverment last years

4. We will be met by her parents at the station tomorrow 

5. A meeting is being held by Mr. Brown in the hall 

Bài 2 

1 Tea can't be made with cold water

2 Some of my books have been taken away.

3 Some pictures were taken away by the boys.

4 This room may be used for the classroom.

5 This machine mustn't be used after 5:30 p.m 

6 Mr Cole used to be visited at weekends by John.

7 All the homework ought to be done by her

bài hình mà không gửi hình, làm bài = niềm tin hả bạn 

Bài 11: 

a) Xét ΔBAD vuông tại A và ΔBED vuông tại E có 

BD chung

\(\widehat{ABD}=\widehat{EBD}\)(BD là tia phân giác của \(\widehat{ABE}\))

Do đó: ΔBAD=ΔBED(Cạnh huyền-góc nhọn)

Suy ra: BA=BE(Hai cạnh tương ứng) và DA=DE(Hai cạnh tương ứng)

Ta có: BA=BE(cmt)

nên B nằm trên đường trung trực của AE(1)

Ta có: DA=DE(cmt)

nên D nằm trên đường trung trực của AE(2)

Từ (1) và (2) suy ra BD là đường trung trực của AE(đpcm)