a)
\(ĐKXĐ:\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\9-x^2\ne0\\2x+1\ne0\end{matrix}\right.< =>\left\{{}\begin{matrix}x\ne-3\\x\ne3\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left|x-2\right|=1=>\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\left(loai\right)\\x=1\end{matrix}\right.\)
\(\left(\dfrac{x-1}{x+3}+\dfrac{2}{x-3}+\dfrac{x^2+3}{9-x^2}\right):\dfrac{-2}{2x+1}\)
\(=\left(\dfrac{x-1}{x+3}+\dfrac{2}{x-3}-\dfrac{x^2+3}{\left(x-3\right)\left(x+3\right)}\right)\cdot\dfrac{2x+1}{-2}\)
\(=\left(\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2+3}{\left(x-3\right)\left(x+3\right)}\right)\cdot\dfrac{2x+1}{-2}\)
\(=\dfrac{x^2-3x-x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{2x+1}{-2}\)
\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{2x+1}{-2}\)
\(=\dfrac{-2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{2x+1}{-2}\)
\(=\dfrac{2x+1}{x+3}\)
với x=1 ta có
\(\dfrac{2\cdot1+1}{1+3}=\dfrac{3}{4}\)
c)
\(\dfrac{2x+1}{x+3}\\ =\dfrac{x+3+x-2}{x+3}\\ =\dfrac{x+3}{x+3}+\dfrac{x-2}{x+3}\\ =1+\dfrac{x-2}{x+3}\\ =1+\dfrac{x+3-5}{x+3}\\ =1+\dfrac{x+3}{x+3}+\dfrac{-5}{x+3}\\ =2+\dfrac{-5}{x+3}\)
để A nhận số nguyên => -5⋮x+3
=> x+3 thuộc ước của -5
\(Ư\left(-5\right)\in\left\{-1;1;-5;5\right\}\)
ta có bảng sau
| x+3 | -1 | 1 | -5 | 5 |
| x | -4(tm | -2(tm | -8(tm | 2(tm |
vậy \(x\in\left\{-4;-2;-8;2\right\}\)
giúp mk vs ạ ai nhanh mk tick nha 
