Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Stress:
1.
A. poem
B. father
C. lucky
D. designer
2.
A. modern
B. novel
C. musician
D. lovely
3.
A. frequent
B. special
C. curly
D. addition
4.
A. public
B. occasion
C. player
D. problem.
5.
A. effect
B. beauty
C. listen
D. brother.
1. For several weeks of preparation beforehand ( about December first ).
2. Children hang up the pillow case or sack.
3. It's reindeer.
4. It starts properly on 24 December.
5. Every year in Britain
b) \(\dfrac{2}{x^2+2x}+\dfrac{8-2x}{x^3+8}=\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x^2-2x+4\right)}\)\(=\dfrac{2\left(x^2-2x+4\right)}{x\left(x+2\right)\left(x^2-2x+4\right)}+\dfrac{2x}{x\left(x+2\right)\left(x^2-2x+4\right)}\)
\(=\dfrac{2x^2-4x+8+2x}{x\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{2x^2-2x+8}{x\left(x+2\right)\left(x^2-2x+4\right)}\)
c) \(\dfrac{x}{1-x^3}+\dfrac{1}{x^2+x-2}=\dfrac{-x}{x^3-1}+\dfrac{1}{x^2+2x-x-2}\)
\(=\dfrac{-x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x\left(x+2\right)-\left(x+2\right)}\)
\(=\dfrac{-x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}\)
\(=\dfrac{-x\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)}+\dfrac{1\left(x^2+x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)}\)\(=\dfrac{-x^2+2x+x^2+x+1}{\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)}\)
f) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)+6x^2-4x+2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{2x-1-6x^2+3x+6x^2-4x+2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{-2x+1}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}=\dfrac{-1}{2x}\)
h) \(\dfrac{1}{2x^2-x-1}+\dfrac{1}{6x^2+9x+3}\)
\(=\dfrac{1}{2x^2+x-2x-1}+\dfrac{1}{3\left(2x^2+3x+1\right)}\)
\(=\dfrac{1}{x\left(2x+1\right)-\left(2x+1\right)}+\dfrac{1}{3\left(2x^2+x+2x+1\right)}\)
\(=\dfrac{1}{\left(2x+1\right)\left(x-1\right)}+\dfrac{1}{3\left[x\left(2x+1\right)+\left(2x+1\right)\right]}\)
\(=\dfrac{1}{\left(2x+1\right)\left(x-1\right)}+\dfrac{1}{3\left(x+1\right)\left(2x+1\right)}\)
\(=\dfrac{3x+3+x-1}{3\left(2x+1\right)\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{4x+2}{3\left(2x+1\right)\left(x+1\right)\left(x-1\right)}=\dfrac{2}{3\left(x+1\right)\left(x-1\right)}\)
e) \(\dfrac{x}{x+2}+\dfrac{4}{x-2}+\dfrac{8x}{4-x^2}\)
\(=\dfrac{x}{x+2}+\dfrac{4}{x-2}+\dfrac{-8x}{x^2-2^2}\)
\(=\dfrac{x}{x+2}+\dfrac{4}{x-2}+\dfrac{-8x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x+4x+8-8x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-6x+8}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x-4x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x-2\right)-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-4}{x+2}\)
That is a peaceful and beautiful countryside. Far away, there are the high mountains. The below mountain is the rice fields.Near the field there are buffaloes eatng grass.At the beginning of the village, there is a big Banyan Tree and the children are play here. In the village there are many trees. Near the river banks, there are a lot of simple houses but very beautiful.
Sẽ gỡ sau 48h.
ecr=Et cetera:vân vân
Đây là 1 meme khá quen thuộc đối với những bạn trẻ ko nghiêm túc. Nó phản ánh khá nhiều về học tập :)), có khá nhiều bn đã sử dụng nó đúng ko nào, nhưng hình ảng chỉ mang tihs chất giải trí, đừng nghĩ xấu về nó :v