\(A=\left(\frac{\left(1-x\right)\left(x-1\right)-\left(x+3\right)^2}{\left(x+3\right)\left(x-1\right)}\right):\left(\frac{\left(x+3\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}\right)\)(ĐKXĐ: x khác -3 và 1)

\(=\frac{-x^2+2x-1-x^2-6x-9}{\left(x+3\right)\left(x-1\right)}:\frac{x^2+6x+9-x^2+2x-1}{\left(x-1\right)\left(x+3\right)}\)

\(=\frac{-2x^2-4x-10}{\left(x+3\right)\left(x-1\right)}:\frac{8x+8}{\left(x-1\right)\left(x+3\right)}\)

\(=\frac{-2x^2-4x-10}{8x+8}\)

Mà \(-2x^2-4x-10=-2\left(x+1\right)^2-8< 0\forall x\)

Nên để A < 0 thì \(8x+8>0\Leftrightarrow x>-1\)

Vậy với \(x>-1,x\ne1\)thì A < 0

\(Giải:\)

\(ĐK:x\ne\left(-2\right);x\ne\left(-1\right)\)

\(\frac{x^2+2x+2}{x+1}>\frac{x^2+4x+5}{x+2}-1\Leftrightarrow\frac{x^2+2x+2}{x+1}>\frac{x^2+3x+3}{x+2}\)

\(\Leftrightarrow\frac{x^2+2x+1}{x+1}+\frac{1}{x+1}-\frac{x^2+3x+2+1}{x+2}>0\)

\(\Leftrightarrow\frac{\left(x+1\right)^2}{x+1}-\frac{\left(x+1\right)\left(x+2\right)}{x+2}+\frac{1}{x+1}-\frac{1}{x+2}>0\)

\(\Leftrightarrow x+1-x-1+\frac{1}{x+1}-\frac{1}{x+2}>0\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}>0\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}=\frac{1}{\left(x+1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x+1>0\\x+2>0\end{cases}}hoặc\hept{\begin{cases}x+1< 0\\x+2< 0\end{cases}}\)

\(+,\hept{\begin{cases}x+1>0\\x+2>0\end{cases}}\Rightarrow x>\left(-2\right)\)

\(+,\hept{\begin{cases}x+1< 0\\x+2< 0\end{cases}}\Rightarrow x< \left(-2\right)\)

BPT đã được giải quyết

BPT <=> -3x²+15x-12>0

<=> x²-5x+4<0

<=> (x-1)(x-4)<0

<=> \(\hept{\begin{cases}x-1>0\\x-4< 0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1< 0\\x-4>0\end{cases}}\)(loại)

<=> 1<x<4

\(\left(x^3-27\right)\left(x^3-1\right)\left(2x+3-x^2\right)\ge0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)\left(x-1\right)\left(x^2+x+1\right)\left[4-\left(x-1\right)^2\right]\ge0\)

\(\Leftrightarrow\left(x-3\right)\left[\left(x+\frac{3}{2}\right)^2+\frac{27}{4}\right]\left(x-1\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\left(4-x+1\right)\left(4+x-1\right)\ge0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(5-x\right)\left(x+3\right)\left[...\right]\left[...\right]\ge0\)(1)

Do [...] và [...] > 0

nên \(\left(1\right)\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(5-x\right)\left(x+3\right)\ge0\)

               \(\Leftrightarrow\left(x-5\right)\left(x-3\right)\left(x-1\right)\left(x+3\right)\le0\)

Có: \(x-5< x-3< x-1< x+3\)

Nên xảy ra các trường hợp sau :

TH1:\(\hept{\begin{cases}x-5\le0\\x-3\ge0\end{cases}}\)(Tự giải)

TH2:\(\hept{\begin{cases}x-1\le0\\x+3\ge0\end{cases}}\)(Tự giải)

Cuối cùng gộp khoảng (Nếu được)

Kết luận......

1/ she had her hair cut

2/ Experiments should be stopped on animal

3/ He wasn't taken to the movie last night

4/ By whom has this wall been rebuilt?

5/ My house needs painting

D O I H A V E T O A N S W E R T H I S ?

Ý nói trong suốt cuộc đời chưa từng được đối xử tốt như vậy

MEAN THAT NO ONE HAVE EVER TREATED HIM SUCH FRIENDLINESS

I love a lot of subjects, but there is the only one subject I the most. It's absolutely English. English helps me a lot. For example, when I go to another countries, I can talk with the foreigns by English, I don't have to worry anything 'bout my language anymore. So, I always spend 2 hours to prace and study this subject at home every evening. And I can learn English at school three time a week, too. The special thing of this subject is there are thousands or may be millions of difficult words I've ever seen before. It makes me feel interesting about English. It's really really great if you know all of the English word, right? But it's also very difficult to perform, so to improve your English vocabulary, there's only one way, patient and keep prace memorize the words. Not lucky, it's the best way! However, I always love this subject because it also helps me to improve my personality - patient!!! 

câu 1
- Because it'sgood for me / hoặc theo ý bạn 

-Yes, i often learn this subject

-i give my subjecct các thứ