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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. 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
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.
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.
- What is this learning technique?
=> That technique is use flashcards to memorize
- Do you often learn this way?
=> Sure, I usually learn by flashcard
- Is it effective with you or not?
=> It is so effective for me
- Do you have any other ways to learn English new words?
=> I can learn new words by some steps such as music, by game,...