Gọi x, y lần lượt là số bút, vở mà bạn Hưng mưa.

Điều kiện: \(x,y\inℕ^∗\)

Theo đề bài, ta có:

+ Tổng số bút, vở là: 14

\(\rightarrow x+y=14\left(1\right)\)

+ Tổng số tiền phải trả là: 100000 đồng, 1 cái bút giá 5000 đồng, 1 quyển vở giá 10000 đồng.

\(\rightarrow5000x+10000y=100000\left(2\right)\Rightarrow x+2y=20000\)

Từ (1) và (2), ta có:

\(\left\{{}\begin{matrix}x+y=14\\x+2y=20000\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=14\\y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=6\end{matrix}\right.\) (không thỏa mãn)

1. I have never talked to a foreigner before.
2. Long was sleepy, but he still stayed up late to watch the end of the game on TV.
3. Living in the countryside is what she likes.
4. Do you want a cup of coffee?
5. You should play sports.
6. You shouldn't play cards all night.
7. How many people are there in Vietnam?

\(\left[19\left(4\cdot2^3+18\right)-9\cdot50\right]:5^2\\ =\left[19\left(4\cdot8+18\right)-450\right]:25\\ =\left[19\cdot50-450\right]:25\\ =\left[950-450\right]:25\\ =500:25\\ =20\)

Ta có:

\(\dfrac{4}{5}\) thế kỉ \(=80\) năm

Cây đo đó trồng năm là:

\(2024-80=1944\left(năm\right)\)

\(x+62^2=x+3844\)

\(\rightarrow\) Đề chưa đầy đủ ạ

\(\dfrac{a+13}{a+11}=\dfrac{a+11+2}{a+11}=1+\dfrac{2}{a+11}\)

\(\dfrac{a+2023}{a+2021}=\dfrac{a+2021+2}{a+2021}=1+\dfrac{2}{a+2021}\)

Vì: \(\dfrac{2}{a+11}>\dfrac{2}{a+2021}\) nên \(\dfrac{a+13}{a+11}>\dfrac{a+2023}{a+2021}\)

\(20,46cm^2=20,46:10000=0,002046m^2\)

Vì: \(1cm^2=\dfrac{1}{10000}m^2=0,0001m^2\)

Sửa đề:

\(\left(7^5+7^9\right)\cdot\left(5^4+5^6\right)\cdot\left(3^3\cdot3-9^2\right)\\ =\left(7^5+7^9\right)\cdot\left(5^4+5^6\right)\cdot\left(3^4-3^4\right)\\ =\left(7^5+7^9\right)\cdot\left(5^4+5^6\right)\cdot0\\ =0\)

Ta có:

\(\dfrac{3}{12}=\dfrac{3}{12}=1\)

\(\dfrac{3}{13}< \dfrac{3}{12}\)

\(\dfrac{3}{14}< \dfrac{3}{12}\)

\(\dfrac{3}{15}< \dfrac{3}{12}\)

\(\dfrac{3}{16}< \dfrac{3}{12}\)

\(\dfrac{3}{17}< \dfrac{3}{12}\)

\(\dfrac{3}{18}< \dfrac{3}{12}\)

\(\rightarrow\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}+\dfrac{3}{15}+\dfrac{3}{16}+\dfrac{3}{17}+\dfrac{3}{18}< \dfrac{3}{12}+\dfrac{3}{12}+\dfrac{3}{12}+\dfrac{3}{12}+\dfrac{3}{12}+\dfrac{3}{12}+\dfrac{3}{12}=\dfrac{3+3+3+3+3+3+3}{12}=\dfrac{21}{12}=\dfrac{7}{4}\)Vậy: \(\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}+\dfrac{3}{15}+\dfrac{3}{16}+\dfrac{3}{17}+\dfrac{3}{18}< \dfrac{7}{4}\)

\(\dfrac{22}{25}-\dfrac{4}{5}=\dfrac{22}{25}-\dfrac{20}{25}=\dfrac{22-20}{25}=\dfrac{2}{25}\)