Em ơi đăng tách bài ra mỗi lượt đăng 1-2 bài thôi nha!

Bài 2 : (1) liên kết ; (2) electron ; (3) liên kết ; (4) : electron ; (5) sắp xếp electron

Bài 4 : 

$\dfrac{M_X}{4} = \dfrac{M_K}{3} \Rightarrow M_X = 52$

Vậy X là crom,KHHH : Cr

Bài 5 : 

$M_X = 3,5M_O = 3,5.16 = 56$ đvC

Tên : Sắt

KHHH : Fe

Bài 9 : 

$M_Z = \dfrac{5,312.10^{-23}}{1,66.10^{-24}} = 32(đvC)$

Vậy Z là lưu huỳnh, KHHH : S

Bài 10  :

a) $PTK = 22M_{H_2} = 22.2 = 44(đvC)$

b) $M_{hợp\ chất} = X + 16.2 = 44 \Rightarrow X = 12$
Vậy X là cacbon, KHHH : C

Bài 11 : 

a) $PTK = 32.5 = 160(đvC)$

b) $M_{hợp\ chất} = 2A + 16.3 = 160 \Rightarrow A = 56$
Vậy A là sắt

c) $\%Fe = \dfrac{56.2}{160}.100\% = 70\%$

7. How often does he go to the library?
8. ....is the shortest student in his class.
9......is her address?
10... is shorter than Ba.

7 how often does he go to the library

8 nga is the shortest student in the class

9 what is her adress

10 hoa is shorter than ba

Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2022^2}< \dfrac{1}{2021.2022}\)

cộng vế với vế 

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2022^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

Vậy ta có đpcm 

mình làm những bài bn chưa lm nhé

9B

10A

bài 2

have repainted

bàii 3

ride - walikking

swimming

watch

Dù sao cũng cảm ơn bạn 🥰

Bài 2:

1: \(x+\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{31\cdot33}=3\)

=>\(x+\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{31\cdot33}\right)=3\)

=>\(x+\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{31}-\frac{1}{33}\right)=3\)

=>\(x+\frac12\left(1-\frac{1}{33}\right)=3\)

=>\(x+\frac12\cdot\frac{32}{33}=3\)

=>\(x+\frac{16}{33}=3\)

=>\(x=3-\frac{16}{33}=\frac{99}{33}-\frac{16}{33}=\frac{83}{33}\)

2: \(x-\frac{3}{1\cdot5}-\frac{3}{5\cdot9}-\cdots-\frac{3}{61\cdot65}=2\)

=>\(x-\frac34\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\cdots+\frac{4}{61\cdot65}\right)=2\)

=>\(x-\frac34\left(1-\frac15+\frac15-\frac19+\cdots+\frac{1}{61}-\frac{1}{65}\right)=2\)

=>\(x-\frac34\left(1-\frac{1}{65}\right)=2\)

=>\(x-\frac34\cdot\frac{64}{65}=2\)

=>\(x-\frac{48}{65}=2\)

=>\(x=2+\frac{48}{65}=\frac{130}{65}+\frac{48}{65}=\frac{178}{65}\)

Bài 1:

2: \(B=\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\cdots+\frac{1}{89\cdot90}\)

\(=\frac16-\frac17+\frac17-\frac18+\cdots+\frac{1}{89}-\frac{1}{90}\)

\(=\frac16-\frac{1}{90}=\frac{14}{90}=\frac{7}{45}\)

3: \(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\cdots+\frac{1}{60\cdot62}\)

\(=\frac12\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\cdots+\frac{2}{60\cdot62}\right)\)

\(=\frac12\left(\frac12-\frac14+\frac14-\frac16+\cdots+\frac{1}{60}-\frac{1}{62}\right)\)

\(=\frac12\left(\frac12-\frac{1}{62}\right)=\frac12\cdot\frac{30}{62}=\frac12\cdot\frac{15}{31}=\frac{15}{62}\)

4: \(D=\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+\cdots+\frac{2}{92\cdot95}\)

\(=\frac23\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\cdots+\frac{3}{92\cdot95}\right)\)

\(=\frac23\left(\frac12-\frac15+\frac15-\frac18+\cdots+\frac{1}{92}-\frac{1}{95}\right)\)

\(=\frac23\left(\frac12-\frac{1}{95}\right)=\frac23\cdot\frac{93}{190}=\frac{1}{95}\cdot31=\frac{31}{95}\)

5:Sửa đề: \(E=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdots+\frac{61}{63\cdot65}\)

\(=3\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{63\cdot65}\right)\)

\(=3\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{63}-\frac{1}{65}\right)\)

\(=3\left(1-\frac{1}{65}\right)=3\cdot\frac{64}{65}=\frac{192}{65}\)

29C

30B

31B

1 D

2 C

3 B

4 A

5 B

6 C

1 D

2 C

3 B

4 A

5 B

6 C