tự viết nha

sao gọi ngữ văn lớp 7

a: ΔABC có BH\(\perp\)AC

nên \(S_{ABC}=\dfrac{1}{2}\cdot BH\cdot AC\)

ΔACB có CK\(\perp\)AB

nên \(S_{ABC}=\dfrac{1}{2}\cdot CK\cdot AB\)

b: \(S_{ABC}=\dfrac{1}{2}\cdot CK\cdot AB\)

\(S_{ABC}=\dfrac{1}{2}\cdot BH\cdot AC\)

Do đó: \(CK\cdot AB=BH\cdot AC\)

mà AB=AC

nên BH/CK=1

c: Vì BH/CK=1

nên BH=CK

HELP ME 

C đúng

nhớ tick nhé

A.

1. How far is it from Vinh to Ha Noi City?

2. I used to walk to school when I was a student.

3. In spite of their shortness, they still love playing sports.

4. Although she is poor, everybody has a great regard for her.

5. Although Tien Dung goalkeeper is so young, he performs excellently.

6. In spite of their old age, they worked very hard.

7. Because of Peter's sickness, he didn't go to school yesterday.

B.

1. How far is it from your house to school?

2. Where did you go with your friends yesterday?

3. What subject did he teach in the high school?

4. Who did you see last night?

A

1 How far is it from Vinh to Ha Noi city?

2 I used to walk to school when I was a student

3 In spite of being short, they still love playing sports

4 Although she was poor, everybody has a great regard for her

6 Althugh Tien Dung goalkeeper was so young, he performs excellently

7 In spite of being old, they worked very hard

8 Because of being sick, Peter didn't go to school yesterday

1. D

2. A

3. B

4. C

5. C

6. A

7. B

8. B

9. C

10. C

1 D

2 A

3 B

4 C

5 C

6 A

7 B

8 B

9 C

10 C

Question V:

1. goes

2. go

3. seeing

4. to play

5. has learning

6. watched

7. will visit

8. working

Question VI:

1. celebration

2. festive

3. exciting

4. amazed

V

1 goes

2 go

3 seeing

4 to play

5 has learned

6 watched

7 is going to visit

8 working

\(A\left(1\right)=a\cdot1^2+b\cdot1+c=a+b+c\)

\(A\left(-3\right)=a\cdot\left(-3\right)^2+b\cdot\left(-3\right)+c=9a-3b+c\)

Ta có:

\(A\left(1\right)+A\left(-3\right)=\left(a+b+c\right)+\left(9a-3b+c\right)\\ =10a-2b+2b\\ =2\left(5a+c\right)-2b\\ =2b-2b=0\\ =>A\left(1\right)+A\left(-3\right)=0=>A\left(1\right)=-A\left(-3\right)\)

Ta có:\(A\left(1\right)\cdot A\left(-3\right)=A\left(-1\right)\cdot\left[-A\left(-1\right)\right]=-\left[A\left(-1\right)\right]^2\le0\)

A(1)=a⋅12+b⋅1+c=a+b+c

�(−3)=�⋅(−3)2+�⋅(−3)+�=9�−3�+�A(−3)=a⋅(−3)2+b⋅(−3)+c=9a−3b+c

Ta có:

$A\left(1\right)+A\left(-3\right)=\left(a+b+c\right)+\left(9a-3b+c\right)=10a-2b+2b=2\left(5a+c\right)-2b=2b-2b=0=>A\left(1\right)+A\left(-3\right)=0=>A\left(1\right)=-A\left(-3\right)$

Ta có:�(1)⋅�(−3)=�(−1)⋅[−�(−1)]=−[�(−1)]2≤0A(1)⋅A(−3)=A(−1)⋅[−A(−1)]=−[A(−1)]

Còn tùy mấy môn khác nữa em

Cái đó thì còn phải trông chờ vào điểm tổng kết các môn khác nữa nhé!