a: ĐKXĐ của A là \(x^2-x-6\ne0\)
=>\(\left(x-3\right)\left(x+2\right)\ne0\)
=>\(x\notin\left\{3;-2\right\}\)
\(x^2-4=0\)
=>\(x^2=4\)
=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Thay x=2 vào A, ta được:
\(A=\dfrac{-3\left(2+1\right)}{2^2-2-6}=\dfrac{-3\cdot3}{4-8}=\dfrac{-9}{-4}=\dfrac{9}{4}\)
b: \(B=\dfrac{2x}{x+3}-\dfrac{x}{3-x}-\dfrac{3x^2+9}{x^2-9}\)
\(=\dfrac{2x}{x+3}+\dfrac{x}{x-3}-\dfrac{3x^2+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x\left(x-3\right)+x\left(x+3\right)-3x^2-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{2x^2-6x+x^2+3x-3x^2-9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{-3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3}{x-3}\)
c: \(P=B:A=\dfrac{-3}{x-3}:\dfrac{-3\left(x+1\right)}{x^2-x-6}\)
\(=\dfrac{-3}{x-3}\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{-3\left(x+1\right)}=\dfrac{x+2}{x+1}\)
Để P nguyên thì \(x+2⋮x+1\)
=>\(x+1+1⋮x+1\)
=>\(1⋮x+1\)
mà x+1>=1(Do x là số tự nhiên nên x>=0)
=>x+1=1
=>x=0(nhận)
