a: 2x-1=7
=>2x=8
=>x=4
Khi x=4 thì \(B=1-\dfrac{1}{4+3}=1-\dfrac{1}{7}=\dfrac{6}{7}\)
b: P=A:B
\(=\dfrac{2x^2+4x+13-3x-9-x^2+3x}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+3-1}{x+3}\)
\(=\dfrac{x^2+4x+4}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+2}=\dfrac{x+2}{x-3}\)
c: Để P là số nguyên thì \(x-3+5⋮x-3\)
=>\(x-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{4;2;8\right\}\)