Bài 1:
a: \(P=\dfrac{x^2+6x+9-\left(x^2-6x+9\right)-4x^2}{\left(3-x\right)\left(3+x\right)}:\dfrac{2x+1-x-3}{3+x}\)
\(=\dfrac{-3x^2+6x+9-x^2+6x-9}{3-x}\cdot\dfrac{1}{x-2}\)
\(=\dfrac{4x^2-12x}{x-3}\cdot\dfrac{1}{x-2}=\dfrac{4x}{x-2}\)
b: \(2x^2-5x+2=0\)
=>(2x-1)(x-2)=0
=>x=1/2
Thay x=1/2 vào P, ta được:\(P=\dfrac{4\cdot\dfrac{1}{2}}{\dfrac{1}{2}-2}=\dfrac{2}{-\dfrac{3}{2}}=-3\)
c: Để P là số nguyên dương thì \(\left\{{}\begin{matrix}4x⋮x-2\\\left[{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\\\left[{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{4;6;-2;10;-6\right\}\)
