a: \(\widehat{B}=\dfrac{180^0-80^0}{2}=50^0\)

b: Chiều dài là \(\sqrt{15^2-9^2}=12\left(dm\right)\)

1 poverty

2 decision

3 worried

4 awake

5 singers

7 northen

8 disappeared

9 difference

10 unimportant

11 windy

12 portable

13 sensibly

14 painting

15 mainly

SOS😥👉🏳👈

Em cần hỗ trợ câu nào em nhỉ?

\(a) 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ b) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ c) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ d) C_4H_{10} + \dfrac{13}{2}O_2 \xrightarrow{t^o} 4CO_2 + 5H_2O\\ e) 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ g) 2KClO_3 \xrightarrow{^o} 2KCl + 3O_2\)

a) \(2Mg+O_2-^{t^o}\rightarrow2MgO\)

b)\(4P+5O_2-^{t^o}\rightarrow2P_2O_5\)

c)\(2KMnO_4-^{t^o}\rightarrow K_2MnO_4+MnO_2+O_2\)

\(2C_4H_{10}+13O_2\rightarrow8CO_2+10H_2O\)

\(3Fe+2O_2\rightarrow Fe_3O_4\)

\(2KClO_3\rightarrow2KCl+3O_2\)

( Làm nốt :vvv )

I

1 C

2 A

3 D

4 C

5 D

II

1 after

2 on

3 didn't use to

4 working

5 because

6 natural

7 don't they

8 whom

9 yelling

10 export

III

1 to having => to have

2 getting => bỏ

3 can go => could go

4 will be turned => would be turned

5 to go => going

IV

1. I didn't spend as much money during the trip as you did.

2. Is this your first time in/visit to Kuala Lumpur?

3. As China has many old castles and pagodas, it is a huge tourist attraction.

4. I wasn't very interested in this place.

5. It is three years since we last went to Spain.

X x ( 6,2 + 3,8 ) = 24,38

X x 10 = 24,38 

X = 2,438

6,2 x X +X x 3,8=43,18-18,8

X x(6,2+3,8)=25

X x 10=25

X =25:10

X=2,5

vậy X = 2,5

mikthaays bạn hương làm tắt quá ạ

Các bạn chỉ cần giúp mik câu c ạ

\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)

\(=\dfrac{x^3-x\left(x+2\right)+2\left(x-2\right)}{x^2-4}\)

\(=\dfrac{x^3-x^2-2x+2x-4}{x^2-4}\)

\(=\dfrac{x^3-x^2-4}{x^2-4}\)

a,\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{\left(x^2-4\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=\dfrac{\left(x-2\right)\left(x+2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow C=x-1\)

b, C=0\(\Rightarrow x-1=0\Rightarrow x=1\)

c, Để C nhận giá trị dương thì \(x-1\ge0\Rightarrow x\ge1\)

\(\frac{2^3.3}{2^23^2.5}=\frac{2}{3.5}=\frac{2}{15}\)

\(\frac{2^3.3}{2^2.3^2.5}=\frac{2}{3.5}=\frac{2}{15}\)

Thiếu dấu nhân ở chỗ \(2^2.3^2\)nha 

cho mik hỏi bạn tùng lâm là bạn hướng dẫn mik cách giải bài này đc ko tại vì mik vẫn còn mấy bài giống vậy