a) \(\frac{3}{4} + \frac{- 1}{3} + \frac{- 5}{18} = \frac{27}{36} + \frac{- 12}{36} + \frac{- 10}{36} = \frac{5}{36}\).

b) \(13 , 57.5 , 5 + 13 , 57.3 , 5 + 13 , 57 = 13 , 57. \left(\right. 5 , 5 + 3 , 5 + 1 \left.\right) = 13 , 57.10 = 135 , 7.\)

Renewable energy sources are important because they are clean, which means they don't harm the environment. They are also sustainable because they won't run out. Additionally, they help reduce pollution and greenhouse gas emissions. Using renewable energy can also create new job opportunities. Overall, renewable energy helps us protect the planet for future generations.

There are some students in the picture. There are many bookshelves around, so I guess they are studying in the library. The boy is carrying a laptop, while the girl is carrying many books. Two friends are discussing with each other. They might be reviewing and preparing for the upcoming exam.

1. My name is …

2. I ride my bicycle every day, usually in the morning or afternoon, for exercise and to get around town easily.

3. I like action and comedy films the most because they are exciting and make me laugh.

4. I joined the Lunar New Year festival because it's a traditional celebration with lots of food, performances, and fireworks.

5. I use electricity and sometimes gas for cooking at home because they are convenient and reliable sources of energy

1. The film was a success despite its low budget / having a low budget. 

2. You mustn’t throw trash out of your car window in traffic. 

3. The food at that stall was more delicious than the food at this one. 

4. There are not enough notebooks for the students. 

1.I think we will travel by hyperloop in the future. 

2.Traffic in big cities is often worse than in small towns.

3. Are there usually special activities for children at festivals?

1. Motorbike taxis, called "xe om," are popular for short trips.

2. People can watch films on streaming services like Netflix.

3. Waterfalls and rivers are natural sources of hydropower.

Gọi số sách lớp 7A; 7B quyên góp được lần lượt là \(x , y\) ( ĐK: \(x , y \in \&\text{nbsp}; \mathit{N}^{*}\))

Theo đề bài:

+) Lớp 7A và 7B quyên góp được \(121\) quyển sách

Nên ta có: \(x + y = 121\)

+) Số sách giáo khoa của lớp 6A; lớp 6B tỉ lệ thuận với tỉ lệ thuận với 5; 6

Nên ta có: \(\frac{x}{5} = \frac{y}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có \(\frac{x}{5} = \frac{y}{6} = \frac{x + y}{5 + 6} = \frac{121}{11} = 11\)

Suy ra: x=55, y= 66 ( thỏa mãn).

Vậy lớp 6A quyên góp được \(55\) quyển sách, lớp 6B quyên góp được \(66\) cuốn.

Tại \(x = 9\) thì:

\(\mathit{C} = x^{14} - 10 x^{13} + 10 x^{12} - 10 x^{11} + . . . + 10 x^{2} - 10 x + 10\)

\(\mathit{C} = x^{14} - \left(\right. x + 1 \left.\right) x^{13} + \left(\right. x + 1 \left.\right) x^{12} - \left(\right. x + 1 \left.\right) x^{11} + . . . + \left(\right. x + 1 \left.\right) x^{2} - \left(\right. x + 1 \left.\right) x + x + 1\)

\(\mathit{C} = x^{14} - x^{14} - x^{13} + x^{13} + x^{12} - x^{12} - x^{11} + . . . + x^{3} + x^{2} - x^{2} - x + x + 1\)

\(\mathit{C} = 1\).

Vậy tại \(x = 9\) thì giá trị của \(\mathit{C}\) bằng \(1\).

AECFBH a) Xét \(\Delta \mathit{A} \mathit{H} \mathit{B}\) và \(\Delta \mathit{A} \mathit{H} \mathit{C}\) có:

\(\mathit{A} \mathit{B} = \mathit{A} \mathit{C}\) (gt);

\(\mathit{A} \mathit{H}\) chung;

\(\mathit{H} \mathit{B} = \mathit{H} \mathit{C}\) (\(\mathit{H}\) là trung điểm của \(\mathit{B} \mathit{C}\));

Suy ra \(\Delta \mathit{A} \mathit{H} \mathit{B} = \Delta \mathit{A} \mathit{H} \mathit{C}\) (c.c.c).

b) Vì \(\Delta \mathit{A} \mathit{H} \mathit{B} = \Delta \mathit{A} \mathit{H} \mathit{C}\) (cmt) suy ra \(\hat{\mathit{A} \mathit{H} \mathit{B}} = \hat{\mathit{A} \mathit{H} \mathit{C}}\) (cặp góc tương ứng).

Mà \(\hat{\mathit{A} \mathit{H} \mathit{B}} + \hat{\mathit{A} \mathit{H} \mathit{C}} = 18 0^{\circ}\) (hai góc kề bù).

Suy ra \(\hat{\mathit{A} \mathit{H} \mathit{B}} = \hat{\mathit{A} \mathit{H} \mathit{C}} = 9 0^{\circ}\).

Vậy \(\mathit{A} \mathit{H} \bot \mathit{B} \mathit{C}\).

c) Vi \(\Delta \mathit{A} \mathit{H} \mathit{B} = \Delta \mathit{A} \mathit{H} \mathit{C}\) (cmt) suy ra \(\hat{\mathit{H} \mathit{A} \mathit{B}} = \hat{\mathit{H} \mathit{A} \mathit{C}} = 4 5^{\circ}\);

\(\hat{\mathit{H} \mathit{C} \mathit{A}} = \hat{\mathit{H} \mathit{B} \mathit{A}} = \frac{18 0^{\circ} - \hat{\mathit{B} \mathit{A} \mathit{C}}}{2} = 4 5^{\circ}\) (cặp góc tương ứng).

Xét \(\Delta \mathit{E} \mathit{B} \mathit{A}\) và \(\Delta \mathit{B} \mathit{F} \mathit{C}\) có:

\(\mathit{A} \mathit{B} = \mathit{C} \mathit{F}\) (gt);

\(\hat{\mathit{B} \mathit{A} \mathit{E}} = \hat{\mathit{B} \mathit{C} \mathit{F}}\) (cùng bù với \(\hat{\mathit{H} \mathit{A} \mathit{B}} = \hat{\mathit{H} \mathit{C} \mathit{A}} = 4 5^{\circ}\));

\(\mathit{E} \mathit{A} = \mathit{B} \mathit{C}\) (gt);

Suy ra \(\Delta \mathit{E} \mathit{B} \mathit{A} = \Delta \mathit{B} \mathit{F} \mathit{C}\) (c.g.c).

Vậy \(\mathit{B} \mathit{E} = \mathit{B} \mathit{F}\) (cặp cạnh tương ứng).