Bài 2: Chọn C

Bài 4: 

a: \(\widehat{C}=180^0-80^0-50^0=50^0\)

Xét ΔABC có \(\widehat{A}=\widehat{C}< \widehat{B}\)

nên BC=AB<AC

b: Xét ΔABC có AB<BC<AC

nên \(\widehat{C}< \widehat{A}< \widehat{B}\)

My dream house is hi tech house by the sea with many sands and water surroundings. There will have 10 large rooms and it will plant many green trees, flowers and have big swimming pool in the garden. In each rooms, it will have a super smart TV, a super smart car, high-tech robot. the super smart TV will help me surf the Internet, send and receive my email, order food from the supermarket and contact my friends on other planets. High-tech robot will clean the floors, cook the meals, wash the clothes, water the flowers and feed the dogs and cats.

I want to build a dream house with a garden where I can grow my fruits and vegetables.

I want a spacious living room with four other rooms which can be used as bedrooms and a study room. One bedroom will be there for my grandparents and the other one for my parents.

My bedroom must have a very comfortable bed where I can relax. It would also be nice to have a PlayStation in my room. My room should be brightly colored, and on the walls, my favorite cartoon characters should be drawn.

For the living room, there must be a giant television where we can watch movies together. I also wish to have a pet cat or a pet dog, so there must be an area for them in my dream house.

Overall, my dream house is a house where my family, along with my pets, and I can live happily.

Bài 2: 

Ta có: \(3n^3+10n^2-5⋮3n+1\)

\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;-3;3\right\}\)

hay \(n\in\left\{0;-1;1\right\}\)

lên mạng tham khảo đi

\(B=3+3^2+...+3^{100}\)

=>\(3B=3^2+3^3+...+3^{101}\)

=>\(3B-B=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)

=>\(2B=3^{101}-3\)

=>\(2B+3=3^{101}\)

=>\(3^n=3^{101}\)

=>n=101

bạn viết đề lại đi khó đọc quá

3 /7...??/

đúng

tất cả 7x5-9 ở mẫu Dương

e viết thế thì ko ai hiểu là đúng rồi

phải có giải thích chứ

Bài 2: 

a: Ta có: ΔABC cân tại A

mà AH là đường cao

nên AH là tia phân giác của góc BAC

b: Xét ΔAMH vuông tại M và ΔANH vuông tại N có

AH chung

\(\widehat{MAH}=\widehat{NAH}\)

Do đó: ΔAMH=ΔANH

Suy ra: AM=AN

hay ΔAMN cân tại A

c: Xét ΔABC có AM/AB=AN/AC

nên MN//BC

d: \(AH^2-AN^2=HN^2\)

\(BH^2-BM^2=MH^2\)

mà HN=MH

nên \(AH^2-AN^2=BH^2-BM^2\)

hay \(AH^2+BM^2=BH^2+AN^2\)