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Trường mink kiểm tra học kì II

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.

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.

\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1=1-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}-\frac{x^2+2}{x^2+2}+\frac{12}{x^2+8}-\frac{x^2+8}{x^2+8}=\frac{x^2+3}{x^2+3}-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}=\frac{x^2-4}{x^2+3}\)

\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}+\frac{-x^2+4}{x^2+3}=0\)

\(\Leftrightarrow\left(-x^2+4\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)

\(\Leftrightarrow-x^2+4=0\left(\text{vì : }\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\right)\)

<=>(2-x)(2+x)=0

<=>x=2 hoặc x=-2

Vậy S={-2;2}