Question

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Answer 1

\(a,B=\dfrac{\dfrac{1}{2}-1}{\dfrac{1}{2}+3}=-\dfrac{1}{2}:\dfrac{7}{2}=-\dfrac{1}{7}\\ b,A=\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x-3}\\ c,P=AB=\dfrac{x+3}{x-3}\cdot\dfrac{x+1}{x+3}=\dfrac{x+1}{x-3}=1+\dfrac{4}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Leftrightarrow x\in\left\{-1;1;2;4;5;7\right\}\left(tm\right)\)