4:

=>\(8\left[15x+11\right]=64\cdot31:2^2=16\cdot31\)

=>15x+11=2*31=62

=>15x=51

=>x=3,4

4: \(=\dfrac{\left(8^9\cdot13+8^8\cdot27\right)}{2^{26}\cdot5}=\dfrac{2^{27}\cdot13+2^{24}\cdot27}{2^{26}\cdot5}\)

\(=2\cdot\dfrac{13}{5}+\dfrac{1}{4}\cdot\dfrac{27}{5}=\dfrac{131}{20}\)

Câu 16:

PTHH: \(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{Cl_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\n_{NaOH}=\dfrac{600\cdot20\%}{40}=3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) 2 chất p/ứ hết

Mặt khác: \(m_{Cl_2}=1,5\cdot71=106,5\left(g\right)\)

\(\Rightarrow m_{nướcjaven}=m_{Cl_2}+m_{ddNaOH}=706,5\left(g\right)\)

Câu 21:

PTHH: \(H_2SO_{4\left(đ\right)}+CaF_2\rightarrow CaSO_4+2HF\)

Ta có: \(n_{HF}=\dfrac{250\cdot40\%}{20}=5\left(kmol\right)\) 

\(\Rightarrow n_{CaF_2}=2,5\left(kmol\right)\) \(\Rightarrow m_{CaF_2}=2,5\cdot78=195\left(kg\right)\) 

:vvv nhiều quá v bn gửi lẻ ra đi

1. fascinated => fascinating

2. I was embarrassed .....

3. bored => boring

4. welcoming => welcomed

5. bored => boring

6. was => bỏ

7. confusing => confused

8. Đ

b.

c.

Ranh giới giữa các câu ghép được đánh dấu bằng:

- Từ thì ở câu 1.

- Dấu phẩy (,) ở câu 2; dấu hai chấm (:) ở câu 3 và hai dấu chấm phẩy (;) ở câu 4.

c nhé bạn

Câu c) bạn nhé

C đúng nhất

1 better

2 more interesting

3 more dangerous

4 friendlier

5 more relaxed

6 busier

7 slower

8 more boring

9 healthier

10 cleaner 

11 more exciting

9: Ta có: \(\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)\)

\(=\left(\sqrt{2}+1\right)^2-3\)

\(=3+2\sqrt{2}-3=2\sqrt{2}\)

10: Ta có: \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{6}-\sqrt{2}}+2\sqrt{10}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{2}\left(\sqrt{3}-1\right)}+2\sqrt{10}\)

\(=\dfrac{\sqrt{10}}{2}+\dfrac{4\sqrt{10}}{2}=\dfrac{5\sqrt{10}}{2}\)

13)\(\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}=\dfrac{\sqrt{10}\left(2+\sqrt{3}\right)-\sqrt{2}\left(2+\sqrt{3}\right)}{2\left(\sqrt{10}-\sqrt{2}\right)}\)\(=\dfrac{\left(\sqrt{10}-\sqrt{2}\right)\left(2+\sqrt{3}\right)}{2\left(\sqrt{10}-\sqrt{2}\right)}=\dfrac{2+\sqrt{3}}{2}\)

14)sai đề? phải là \(\sqrt{3-\sqrt{5}}\) 

\(=\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{2\sqrt{10}-2\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{10}-2\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{4\left(\sqrt{5}-1\right)}=\dfrac{\left|\sqrt{5}-1\right|\left(3+\sqrt{5}\right)}{4\left(\sqrt{5}-1\right)}\)

\(=\dfrac{3+\sqrt{5}}{4}\)

15)\(\sqrt{\left(1-\sqrt{2016}\right)^2}.\sqrt{2017+2\sqrt{2016}}=\left|1-\sqrt{2016}\right|\sqrt{1+2\sqrt{2016}+2016}\)

\(=\left(\sqrt{2016}-1\right)\sqrt{\left(1+\sqrt{2016}\right)^2}=\left(\sqrt{2016}-1\right)\left(1+\sqrt{2016}\right)\)

\(=2015\)