Part 1

1 On

2 did

3 would

4 therefore

5 who

6 of

Part 2

1 G

2 A

3 F

4 D

5 B

6 C

Part 3

1 are watered

2 will become

3 has read

4 revising

5 communicate

6 announcement

7 interactive

8 conservationist

Part 4 

1 to travel => travelling

2 that => bỏ

Part 5

1 spoke English as

2 is being used by 

3 he was leaving

4 you can find are 

Part 6

1 walking

2 good

3 experienced

4 burden

1  A

2 D

3B

4 C

5 B

Em cần mọi người hỗ trợ những câu nào hay toàn đề em nhỉ? Hay em đăng lên cho các bạn tham khảo đề nè!

a/ Tam giác AMN cân tại A (gt). \(\Rightarrow\) \(\widehat{AMN}=\widehat{ANM};AM=AN.\)

Xét tam giác AMB và tam giác ANC có:

+ AM = AN (cmt).

+ \(\widehat{AMB}=\widehat{ANC}\left(\widehat{AMN}=\widehat{ANM}\right).\)

+ MB = NC (gt).

\(\Rightarrow\) Tam giác AMB = Tam giác ANC (c - g - c).

\(\Rightarrow\) AB = AC (cặp cạnh tương ứng).

Xét tam giác ABC có: AB = AC (cmt).

\(\Rightarrow\) Tam giác ABC cân tại A.

b/ Tam giác ABC cân tại A (cmt) \(\Rightarrow\) \(\widehat{ABC}=\widehat{ACB}.\)

Mà \(\widehat{ABC}=\widehat{MBH;}\widehat{ACB}=\widehat{NCK}\text{​​}\) (đối đỉnh).

\(\Rightarrow\) \(\widehat{MBH}=\widehat{NCK}.\)

Xét tam giác MBH và tam giác NCK \(\left(\widehat{BHM}=\widehat{CKN}=90^o\right)\)có:

+ MB = NC (gt).

+ \(\widehat{MBH}=\widehat{NCK}\left(cmt\right).\)

\(\Rightarrow\) Tam giác MBH = Tam giác NCK (cạnh huyền - góc nhọn).

c/ Tam giác MBH = Tam giác NCK (cmt).

\(\Rightarrow\) \(\widehat{BMH}=\widehat{CNK}\) (cặp góc tương ứng).

Xét tam giác OMN có: \(\widehat{NMO}=\widehat{MNO}\) (do \(\widehat{BMH}=\widehat{CNK}\)).

\(\Rightarrow\) Tam giác OMN tại O.

part 1

1 which

2 won't

3 because

4 Good idea

5 on

6 therefore

part 2

1D 2F 3A 4B 5C 6E

part 3

1 studied

2 has worked

3 being interviewed

4 to be taken

5 would visit

6 put

7 endangered

8 compulsorily

part 4

1 are -> is

2 me -> to me

3 work -> working

4 when -> since

part 5 

1 written novels for

2 to read bedtime stories

3 to wearing

4 has been absent from 

5 doesn't have a passport

6 were given to Martha

7 play football that

8 whom my teacher was

part 6

1 mankind

2 polluted

3 of

4 naturally

part 7

1F 2F 3T 4T

Kẻ AH⊥BC

ta có: \(VP=AB^2+BC^2-2.AB.BC.cosB=AB^2+BC^2-2.AB.BC.\dfrac{BH}{AB}=AB^2+BC^2-2.BH.BC=AB^2-BH^2+BC^2-2.BH.BC+BH^2=AH^2+\left(BC-BH\right)^2=AH^2+CH^2=AC^2=VT\)

8) \(\dfrac{x+7}{3}+\dfrac{x+5}{4}=\dfrac{x+3}{5}+\dfrac{x+1}{6}\)

\(\Rightarrow\dfrac{x+7}{3}+\dfrac{x+5}{4}-\dfrac{x+3}{5}-\dfrac{x+1}{6}=0\)

\(\Rightarrow\dfrac{x+7}{3}+2+\dfrac{x+5}{4}+2-\dfrac{x+3}{5}-2-\dfrac{x+1}{6}-2=0+2+2-2-2\)

\(\Rightarrow\left(\dfrac{x+7}{3}+2\right)+\left(\dfrac{x+5}{4}+2\right)-\left(\dfrac{x+3}{5}+2\right)-\left(\dfrac{x+1}{6}+2\right)=0\)

\(\Rightarrow\left(\dfrac{x+7}{3}+\dfrac{6}{3}\right)+\left(\dfrac{x+5}{4}+\dfrac{8}{4}\right)-\left(\dfrac{x+3}{5}+\dfrac{10}{5}\right)-\left(\dfrac{x+1}{6}+\dfrac{12}{2}\right)=0\)

\(\Rightarrow\left(x+13\right)\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+13=0\\\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=0\end{matrix}\right.\)

\(x+13=0\)

\(\Rightarrow x=-13\)

\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=0\)

\(\dfrac{13}{60}=0\) (vô lí)

Vậy \(x=-13\)

9) Bạn chuyển vế rồi cộng 3 vào từng mỗi số

8) Ta có: \(\dfrac{x+7}{3}+\dfrac{x+5}{4}=\dfrac{x+3}{5}+\dfrac{x+1}{6}\)

\(\Leftrightarrow\dfrac{x+13}{6}+\dfrac{x+13}{4}=\dfrac{x+13}{5}+\dfrac{x+13}{6}\)

Suy ra: x+13=0

hay x=-13

9) Ta có: \(\dfrac{x+19}{3}+\dfrac{x+13}{5}=\dfrac{x+7}{7}+\dfrac{x+1}{9}\)

\(\Leftrightarrow\dfrac{x+28}{3}+\dfrac{x+28}{5}-\dfrac{x+28}{7}-\dfrac{x+28}{9}=0\)

Suy ra: x+28=0

hay x=-28

Bài 8 : 

200ml = 0,2l

\(n_{HCl}=0,2.1=0,2\left(mol\right)\)

Pt : \(R+2HCl\rightarrow RCl_2+H_2|\)

        1       2             1          1

       0,1    0,2                      0,1

a) \(n_{H2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

b) \(n_R=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(M_R=\dfrac{2,4}{0,1}=24\) (g/mol)

Vậy kim loại R là magie

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