mấy bạn viết jum mink vs ( thi nói SPEAKING ) LỚP 8
Topic: A VACATION ABORD
Help me!!! please!!!!
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D
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1 tháng 5 2016
Talk Show ở Trung tâm Better English hả bạn
SM
2
HA
25 tháng 12 2016
Reading books :
I'm very happy to introduce somethings about the activity I do in my free time. I often read books in my free time. I started reading books when I was at the age of six. You know that, I have known how to read books. I read whatever I saw and whenever I had free time. I often went to library or bookshop to find some interesting books and I bought it. Now, I have a large colection of books. I often find out the most interesting of the books give us and share to my sister. I found it very useful. I love reading books a lot and I hope, I will keep doing this activity in the future.
S
_silverlining
CTVVIP
25 tháng 12 2016
There are three activities I like to do in my free time. The first is cartoon. I like watch cartoons very much. I love “Total Spies” most because it is very exciting and funny. I often watch it with my friends to relax. Second, I also play football with my brother and my friends when I have free time. I often play it in our school yard. I think playing football is sport which helps us have good health and study better. Final, I like reading books best. Since reading books are good, you can learn about another culture and again your knowledge. Moreover, it is a form of entertainment, which is useful to our life. All in all, I find that if I haven’t anything to do in my free time, I will be boring. Therefore, you should also choose some activities for yourself to do, which helps you have healthy way of life.
MT
15 tháng 1 2016
\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1=1-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{6}{x^2+2}-\frac{x^2+2}{x^2+2}+\frac{12}{x^2+8}-\frac{x^2+8}{x^2+8}=\frac{x^2+3}{x^2+3}-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}=\frac{x^2-4}{x^2+3}\)
\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}+\frac{-x^2+4}{x^2+3}=0\)
\(\Leftrightarrow\left(-x^2+4\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)
\(\Leftrightarrow-x^2+4=0\left(\text{vì : }\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\right)\)
<=>(2-x)(2+x)=0
<=>x=2 hoặc x=-2
Vậy S={-2;2}
Mình trả lơì cho vì mình học lớp 11 mà
ý bạn là sao vậy .hong hiểu