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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) \(\dfrac{x}{x+2}+\dfrac{4}{x-2}+\dfrac{8x}{4-x^2}\)
\(=\dfrac{x}{x+2}+\dfrac{4}{x-2}+\dfrac{-8x}{x^2-2^2}\)
\(=\dfrac{x}{x+2}+\dfrac{4}{x-2}+\dfrac{-8x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x+4x+8-8x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-6x+8}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x-4x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x-2\right)-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-4}{x+2}\)
1, C, look+adj
2,at, at thường được dùng cho những địa điểm đã được xác định hay những địa đoeerm gần và nhỏ
3,A, join:tham gia, nếu chọn B thì phải có giới từ đi kèm
4,C
5,C, quyên góp tiền
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.
bạn dùng quá khứ đơn á =]]