Bài 1:

a: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

b: P<1/2

=>P-1/2<0

=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)

=>\(\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)

=>\(\sqrt{x}-3< 0\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< 9\\x< >1\end{matrix}\right.\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_S=\dfrac{0,8}{32}=0,025\left(mol\right)\)

PTHH: Fe + S --to--> FeS

LTL: \(0,1>0,025\rightarrow\) Fe dư

Theo pthh: \(n_{Fe\left(pu\right)}=n_{FeS}=n_S=0,025\left(mol\right)\)

\(\rightarrow\left\{{}\begin{matrix}m_{Fe\left(du\right)}=\left(0,1-0,025\right).56=4,2\left(g\right)\\m_{FeS}=0,025.88=2,2\left(g\right)\end{matrix}\right.\)

PTHH:

Fe + 2HCl ---> FeCl₂ + H₂

0,075                          0,075

FeS + 2HCl ---> FeCl₂ + H₂S

0,025                             0,025

\(\rightarrow\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,075}{0,075+0,025}=75\%\\\%V_{H_2S}=100\%-75\%=25\%\end{matrix}\right.\)

- Trích một ít các dd làm mẫu thử, đánh số thứ tự lần lượt

- Cho các dd tác dụng với giấy quỳ tím:

+ QT chuyển đỏ: AgNO₃

+ QT chuyển xanh: Na₂S, Na₂CO₃ (1)

+ QT không chuyển màu: K₂SO₄, KCl (2)

- Cho các dd ở (1) tác dụng với dd HCl dư

+ Có khí mùi trứng thối thoát ra: Na₂S

Na₂S + 2HCl --> 2NaCl + H₂S

+ Có khí không mùi thoát ra: Na₂CO₃

Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

- Cho các dd ở (2) tác dụng với dd BaCl₂

+ Không hiện tượng: KCl

+ Kết tủa trắng: K₂SO₄

\(K_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2KCl\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{H_2SO_4}=\dfrac{62.60\%}{98}=0,38\left(mol\right)\)

PTHH:

Mg + H₂SO₄ ---> MgSO₄ + H₂
a           a                 a           a

Zn + H₂SO₄ ---. ZnSO₄ + H₂

b          b              b            b 

hệ pt \(\left\{{}\begin{matrix}24a+65b=15,4\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Zn}=0,2.65=13\left(g\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{15,4}=15,58\%\\\%m_{Zn}=100\%-15,58\%=84,42\%\end{matrix}\right.\)

b, \(\left\{{}\begin{matrix}m_{MgSO_4}=0,1.120=12\left(g\right)\\m_{ZnSO_4}=0,2.161=32,2\left(g\right)\\m_{dd}=62+15,4-0,3.2=76,8\left(g\right)\\m_{H_2SO_4\left(dư\right)}=\left(0,38-0,1-0,2\right).98=7,84\left(g\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{12}{76,8}=15,625\%\\C\%_{ZnSO_4}=\dfrac{32,2}{76,8}=41,2\%\\C\%_{H_2SO_4}=10,2\%\end{matrix}\right.\)

1. English is more interesting than music.

2. Today they are not as happy as they were yesterday.

3. Ha Noi is not as small as Hai Duong.

4. Mai's sister is not as pretty as her.

6. You have got more money than me.

7. Art is not as difficult as French.

8. Nam's father is more careful than him.

9. No one in our town is as rich as Mr Ron.

10. He is the most intelligent in my class.

11. Everest is the highest mountain in the world.

12. Minh is the fattest person in my group.

13. I can't swim as far as Jan.

14B 15C 16A 17C 18B 19C 20B

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Bài 13:

góc A=180-80-30=70 độ

=>góc BAD=góc CAD=70/2=35 độ

góc ADC=80+35=115 độ

góc ADB=180-115=65 độ

Bài 14: 
Xét ΔABC vuông tại A 
-> \(\widehat{B}\)\(+ \widehat{C}=90^o\)

Mà \(\widehat{B}=\widehat{C}\)
=> \(2\widehat{B}=90^o\)
=> \(\widehat{B}=45^o\)