Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
Thai Nguyen xuan
Xem chi tiết
123211231312
Xem chi tiết
Phía sau một cô gái
10 tháng 10 2023 lúc 20:01

       Christmas is a magical and enchanting time of year. It is a joyous and festive occasion filled with love, happiness, and excitement. The air is filled with a warm and cozy atmosphere, as families come together to celebrate. The beautifully decorated Christmas tree sparkles with twinkling lights and shimmering ornaments. Delicious and indulgent feasts are prepared, filling the air with mouth-watering aromas. Generosity and kindness abound as people exchange thoughtful and heartfelt gifts. The joyful laughter of children fills the air, accompanied by the melodic sounds of Christmas carols. The winter scenery is breathtaking, with glistening snowflakes and frosty landscapes. The spirit of Christmas is truly magical, bringing warmth, joy, and togetherness to all.

Nguyễn Thanh Thủy
21 tháng 6 lúc 21:04

    Christmas is a magical and enchanting time of year. It is a joyous and festive occasion filled with love, happiness, and excitement. The air is filled with a warm and cozy atmosphere, as families come together to celebrate. The beautifully decorated Christmas tree sparkles with twinkling lights and shimmering ornaments. Delicious and indulgent feasts are prepared, filling the air with mouth-watering aromas. Generosity and kindness abound as people exchange thoughtful and heartfelt gifts. The joyful laughter of children fills the air, accompanied by the melodic sounds of Christmas carols. The winter scenery is breathtaking, with glistening snowflakes and frosty landscapes. The spirit of Christmas is truly magical, bringing warmth, joy, and togetherness to all.

Mina Anh
Xem chi tiết
Nguyễn Lê Phước Thịnh
23 tháng 12 2021 lúc 11:43

g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

Nguyễn Hoàng Minh
23 tháng 12 2021 lúc 11:44

\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

Vũ Mai Loan
Xem chi tiết
Trần Đức Huy
6 tháng 2 2022 lúc 11:02

a)\(-1,6:\left(1+\dfrac{2}{3}\right)=-1,6:\dfrac{5}{3}=-\dfrac{8}{5}.\dfrac{3}{5}=\dfrac{-24}{25}\)

b)\(\left(\dfrac{-2}{3}\right)+\dfrac{3}{4}-\left(-\dfrac{1}{6}\right)+\left(\dfrac{-2}{5}\right)=-\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{6}-\dfrac{2}{5}=\dfrac{-40+45+10-24}{60}=\dfrac{-9}{60}=\dfrac{-3}{20}\)

c)\(\left(\dfrac{-3}{7}:\dfrac{2}{11}+\dfrac{-4}{7}:\dfrac{2}{11}\right).\dfrac{7}{33}=\left(\dfrac{-3}{7}.\dfrac{11}{2}+\dfrac{-4}{7}.\dfrac{11}{2}\right).\dfrac{7}{33}=\left[\dfrac{11}{2}\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)\right].\dfrac{7}{33}=\dfrac{-11}{2}.\dfrac{7}{33}=\dfrac{-7}{6}\)

d)\(\dfrac{-5}{8}+\dfrac{4}{9}:\left(\dfrac{-2}{3}\right)-\dfrac{7}{20}.\left(\dfrac{-5}{14}\right)=\dfrac{-5}{8}-\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{8}=\dfrac{-5}{8}+\dfrac{1}{8}-\dfrac{2}{3}=-\dfrac{7}{6}\)

Đặng Minh Dương
Xem chi tiết
Nguyễn Hoàng Minh
8 tháng 11 2021 lúc 9:40

Bài 1:

\(a,\dfrac{25}{14x^2y}=\dfrac{75y^4}{42x^2y^5};\dfrac{14}{21xy^5}=\dfrac{28x}{42x^2y^5}\\ b,\dfrac{3x+1}{12xy^4}=\dfrac{3x\left(3x+1\right)}{36x^2y^4};\dfrac{y-2}{9x^2y^3}=\dfrac{4y\left(y-2\right)}{36x^2y^4}\\ c,\dfrac{1}{6x^3y^2}=\dfrac{6y^2}{36x^3y^4};\dfrac{x+1}{9x^2y^4}=\dfrac{4x\left(x+1\right)}{36x^3y^4};\dfrac{x-1}{4xy^3}=\dfrac{9x^2y\left(x-1\right)}{36x^3y^4}\\ d,\dfrac{3+2x}{10x^4y}=\dfrac{12y^4\left(3+2x\right)}{120x^4y^5};\dfrac{5}{8x^2y^2}=\dfrac{75x^2y^3}{120x^4y^5};\dfrac{2}{3xy^5}=\dfrac{80x^3}{120x^4y^5}\)

Linh nguyễn
Xem chi tiết
Nguyễn Việt Lâm
13 tháng 1 lúc 14:14

\(log_x\left(x^2y^3\right)=log_xx^2+log_xy^3=2+3log_xy\)

\(\Rightarrow2+3log_xy=1\Rightarrow log_xy=-\dfrac{1}{3}\)

\(N=\dfrac{log_x\left(x^2y^3\right)}{log_x\left(\dfrac{\sqrt[5]{x^3y^2}}{xy^3}\right)}=\dfrac{1}{log_x\left(\sqrt[5]{x^3y^2}\right)-log_xxy^3}=\dfrac{1}{log_x\sqrt[5]{x^3}+log_x\sqrt[5]{y^2}-\left(log_xx+log_xy^3\right)}\)

\(=\dfrac{1}{\dfrac{3}{5}+\dfrac{2}{5}log_xy-\left(1+3log_xy\right)}=\dfrac{1}{\dfrac{3}{5}+\dfrac{2}{5}.\left(-\dfrac{1}{3}\right)-1-3.\left(-\dfrac{1}{3}\right)}=\dfrac{15}{7}\)

Thảnh TẠ
Xem chi tiết
Diệp Anh Hạ
Xem chi tiết
Hakimiru Mesuki
26 tháng 3 2023 lúc 16:51

5A

1B

3C

4D

\(#TyHM\)

Pháttài
Xem chi tiết
Lê Phạm Bảo Linh
3 tháng 4 2023 lúc 22:47

1. see off (tiễn)
2. Take off (bỏ ra)
3. go off( xuống xe, rời đi)
4. turn off( tắt)
5. cut off ( rơi ra?)
6. go off( vang lên)
7. get off( ngừng..)
8. go off (ôi thiu)
Mình không thấy đề yêu cầu chia động từ????

Bagel
3 tháng 4 2023 lúc 22:45
Đặng Nguyễn Minh Anh
Xem chi tiết
3e	Nguyễn Ngọc Khánh Lin...
30 tháng 10 2023 lúc 20:13

- Nửa cầu Bắc nằm phía trên đường xích đạo.

 

- Nửa cầu Nam nằm phía dưới đường xích đạo.

 

- Các vĩ tuyến Bắc là các vĩ tuyến nằm từ xích đạo đến cực Bắc.

 

- Các vĩ tuyến Nam là các vĩ tuyến nằm từ xích đạo đến cực Nam.