Truyện trung đại là loại truyện nhiều khi gần với thể kí( Ghi
chép sự việc)với sử( Ghi chép chuyện thật) và có thể hư cấu
thường mang tính giáo huấn. Cốt truyện nhìn chung còn đơn
giản. Nhân vật thường được miêu tả qua hànhđộng và ngôn
ngữ.

1 A

2 C

3 B

4 D

5 A

6 C

7 D

8 D

1 A

2 C

3 B

4 D

5 A

Bài 2: 

Áp dụng định lí Pytago vào ΔEFG vuông tại E, ta được:

\(FG^2=EF^2+EG^2\)

\(\Leftrightarrow FG^2=15^2+5^2=250\)

hay \(FG=5\sqrt{10}\left(cm\right)\)

1 were

2 was 

3 Were

4 was

5 was

6 was 

7 was

B

1 John could play the piano when he was 6

2 My brother could draw beautiful picture when he was seven

3 We could speak 2 foreign languages when we were 18

4 She could dance when she was 8

5 Tiger Woods could play goft when he was 3

C

2 Could you write you name when you were 3

3 Could you read when you were 2

4 Could you count from 1 to 10 when you were 2

5 Could you talk when you were 3

6 Could you ride a bike when you were 6

7 Could you swim when you were 5

A.

1. were

2. was

3. weren't

4. was

5. was

6. was

7. wasn't

B.

2. John could play the piano when he was six.

3. My brother could draw beautiful pictures when he was seven.

4. We could speak two foreign languages when we were eighteen.

5. She could dance when she was eight.

6. Tiger Woods could play golf when he was three.

C.

2. Could you write your name when you were three?

3. Could you read when you were two?

4. Could you count from one to ten when you were two?

5. Could you talk when your were three?

6. Could you ride a bike when you were six?

7. Could you swim when you were five?

Bài 3: 

a, PTHH: Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

b, \(n_{Mg}=\dfrac{7,2}{24}=0,3mol\)

\(n_{MgCl_2}=0,3.1=0,3mol\\ \Rightarrow m_{MgCl_2}=0,3.95=28,5g\)

c, \(n_{H_2}=0,3.1=0,3mol\\ \Rightarrow V_{H_2}=0,3.22,4=6,72l\)

Bài 4: 

a, PTHH: Zn + 2HCl \(\rightarrow\) ZnCl₂ + H₂

b,

 \(n_{Zn}=\dfrac{13}{65}=0,2mol\\ \Rightarrow n_{ZnCl_2}=0,2.1=0,2mol\\ m_{ZnCl_2}=0,2.136=27,2g\)

c, \(n_{H_2}=0,2.1=0,2mol\\ \Rightarrow V_{H_2}=0,2.22,4=4,48l\)

\(\left(2x-3\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\) (Thêm KL cuối dòng: Vậy \(x\in\left\{\dfrac{3}{2};\dfrac{1}{2}\right\}\))

(2x-3)x(x-1/2)=0

Đặt từng nhân tử bằng không và giải cho x:

2x - 3 = 0

2x = 3

x = 3/2

x = 0

x - 1/2 = 0

x = 1/2