a,\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{x+3}{x^2+3x+2x+6}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}+\dfrac{-3x}{3\left(x^2-4\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}+\dfrac{-x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{\left(x+2\right)}:\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{-x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=1+\dfrac{1}{x+2}:\dfrac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{1}{x+2}:\dfrac{6}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{1}{x+2}.\dfrac{\left(x-2\right)\left(x+2\right)}{6}=\dfrac{x-2}{6}\)
b, Để P = 0 ⇔ \(\dfrac{x-2}{6}=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Để \(P=1\Leftrightarrow\dfrac{x-2}{6}=1\Leftrightarrow x-2=6\Leftrightarrow x=8\)
c, Để P > 0 \(\Leftrightarrow\dfrac{x-2}{6}>0\Leftrightarrow x-2>6\Leftrightarrow x>8\)
