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.

    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.

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

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

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

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

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

Bài 1:

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

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

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

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

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

E cảm ơn thầy nhiều ạ 

5A

1B

3C

4D

\(#TyHM\)

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????

- 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.