a: Xét tứ giác ABQN có

\(\widehat{BQN}=\widehat{QNA}=\widehat{NAB}=90^0\)

=>ABQN là hình chữ nhật

b: Xét ΔCAD có

DN,CH là các đường cao

DN cắt CH tại M

Do đó: M là trực tâm của ΔCAD

=>AM\(\perp\)CD

c: Xét ΔHAB vuông tại H và ΔHCA vuông tại H có

\(\widehat{HAB}=\widehat{HCA}\left(=90^0-\widehat{ABC}\right)\)

Do đó: ΔHAB đồng dạng với ΔHCA

=>\(\dfrac{HA}{HC}=\dfrac{HB}{HA}\)

=>\(HA^2=HB\cdot HC\)

=>\(HA=\sqrt{HB\cdot HC}\)

1. Turn on (please+Vo)

2. Turned / off (was chia qk)

3. Look for

4. Got up (because S+ had+V3/ed, S+V2/ed)

5. Ran into( hai hd lien tiep)

6. Goes on (after S+V2/ed, S+Vht)

1. However => Therefore

2. so => but

3. and => but

4. and => but

5. Because => Therefore

6. or => because

7. or => but

8. and => or

9. but => or

10. so => because

Exercise 3:

1. Yes, she is. (Cô ấy bị tàn tật.)

2. They are cleaning the street. (Họ đang dọn dẹp đường phố.)

Phản hồi đến em!

my favourite food is pizza

my favourite drink is Water

i often have noodle for my breakfast

i often have rice , fish , soup , chicken for my lunch

i often rice , meat , soup for my dinner

1. My favourite food is pizza.

2. My favourite drink is water.

3. I often have noodle for my breakfast.

4. I often have rice, fish, soup, chicken for my lunch.

5. I often rice, meat,soup for my dinner.

c)\(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)....\left(1+\dfrac{1}{2020}\right)\left(1+\dfrac{1}{2021}\right)\)

\(=\left(\dfrac{1.2}{1.2}+\dfrac{1}{2}\right)\left(\dfrac{1.3}{1.3}+\dfrac{1}{3}\right)...\left(\dfrac{1.2021}{1.2021}+\dfrac{1}{2021}\right)\)

\(=\dfrac{3}{1.2}\cdot\dfrac{4}{1.3}\cdot\cdot\cdot\cdot\dfrac{2022}{1.2021}\)

\(=\dfrac{3.4.5...2022}{\left(1.1.1....1\right)\left(2.3.4...2021\right)}\)

\(=\)\(\dfrac{3.4.5...2022}{2.3.4...2021}\)

\(=\dfrac{2022}{2}=1011\)

\(d\))\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{199}\right)\left(1-\dfrac{1}{200}\right)\)

\(=\left(\dfrac{2}{1.2}-\dfrac{1}{1.2}\right)\left(\dfrac{3}{1.3}-\dfrac{1}{1.3}\right)....\left(\dfrac{200}{1.200}-\dfrac{1}{1.200}\right)\)

\(=\dfrac{1.2.3....199}{\left(1.1.1....1\right).\left(2.3.4....200\right)}\)

\(=\dfrac{1.2.3...199}{2.3.4...200}\)

Nếu mik làm sai mong bạn thông cảm

1/200

Bạn tham khảo, có j sai thì báo lại mình nhé