\(9,=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2}{4}=\dfrac{1}{2}\\ 10,=\dfrac{\sqrt{5}+2+\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=2\sqrt{5}\\ 11,=\dfrac{8+6\sqrt{2}-8+6\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{12\sqrt{2}}{-2}=-6\sqrt{2}\\ 12,=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}=\dfrac{4}{-2}=-2\\ 13,=\sqrt{2}-1+\sqrt{2}+3=2\sqrt{2}+2\\ 14,=2-\sqrt{3}+\sqrt{3}-1=1\\ 15,=3-\sqrt{5}+\sqrt{5}-2=1\)

1/

1. although

2. around

3. where

4. forward

5. and

6. will

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

3/

1. daily

2. wearing

3. would do

4. haven't eaten

5. is being carried

6. rises

7. inexperienced

8. equality

4/

1. which -> who

2. pollute -> polluting

5/

1. not to touch that

2. Tom has met is

3. weren't ill

4. may be built

6/

1. environmentalists

2. affect

3. media

4. responsible

7/

1. T

2. T

3. F

4. F

Part 6 

1/ students

2/affect

3/media

4/make

khoanh toàn C

1 A

2 D

3 C

4 A

5 C

6 câu này sai 2 chỗ A và C

7 A

8 B

9 B

1 A

2 D

3 C

4 A

5 C

6 A

7 A

8 B

không hỏi bài kiểm tra nha bạn

giúp mk với

Câu 3: 

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{a-c}{9-7}=5\)

Do đó: a=45; b=40; c=35

b: Xét ΔAHB vuông tại H có HE là đường cao ứng với cạnh huyền AB

nên \(AE\cdot AB=AH^2\left(1\right)\)

Xét ΔAHC vuông tại H có HF là đường cao ứng với cạnh huyền AC

nên \(AF\cdot AC=AH^2\left(2\right)\)

Từ (1) và (2) suy ra \(AE\cdot AB=AF\cdot AC\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x-3\\3x+2=3-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-5\\7x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{7}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2+3x=4x-3\\2+3x=3-4x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{7}\end{matrix}\right.\)

a, \(2\sqrt{3}-\sqrt{4+x^2}=0\Leftrightarrow\sqrt{4+x^2}=2\sqrt{3}\)

\(\Leftrightarrow x^2+4=12\Leftrightarrow x^2=8\Leftrightarrow x=\pm2\sqrt{2}\)

b, \(\sqrt{16x+16}-\sqrt{9x+9}=0\)ĐK : x >= -1 

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=0\Leftrightarrow\sqrt{x+1}=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

c, \(\sqrt{4\left(x+2\right)^2}=8\Leftrightarrow2\left|x+2\right|=8\Leftrightarrow\left|x+2\right|=4\)

TH1 : \(x+2=4\Leftrightarrow x=2\)

TH2 : \(x+2=-4\Leftrightarrow x=-6\)

c: Ta có: \(\sqrt{4\left(x+2\right)^2}=8\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

a) PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)

b) Ta có: \(n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{O_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,2\cdot22,4=4,48\left(l\right)\)

c) Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{NaOH}=\dfrac{40\cdot25\%}{40}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo muối trung hòa

PTHH: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

Vì NaOH còn dư nên tính theo CO₂ 

\(\Rightarrow n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\) \(\Rightarrow m_{Na_2CO_3}=0,1\cdot106=10,6\left(g\right)\)