a: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(B=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{9-3x^2}{9-x^2}\)
\(=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x\left(x-3\right)+2x\left(x+3\right)-3x^2+9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2-3x+2x^2+6x-3x^2+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{3}{x-3}\)
b: Để B>0 thì \(\dfrac{3}{x-3}>0\)
=>x-3>0
=>x>3
c: |2x+1|=5
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Thay x=2 vào B, ta được:
\(B=\dfrac{3}{2-3}=\dfrac{3}{-1}=-3\)
d: Để B là số nguyên thì \(3⋮x-3\)
=>\(x-3\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{4;2;6;0\right\}\)
