a/
\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{3x}{3\left(x^2-4\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{x+2}:\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=1+\dfrac{1}{x+2}:\dfrac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{1}{x+2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{6}=1+\dfrac{x-2}{6}\)
\(=\dfrac{6}{6}+\dfrac{x-2}{6}=\dfrac{x+4}{6}\)
b/ +) \(P=0\Leftrightarrow\dfrac{x+4}{6}=0\Leftrightarrow x+4=0\Leftrightarrow x=-4\) (tm)
Vậy x = -4 thì P = 0
+) \(P=1\Leftrightarrow\dfrac{x+4}{6}=1\Leftrightarrow x+4=6\Leftrightarrow x=2\) (ktm)
K có gt nào của x tm P = 1
c/ \(P>0\Leftrightarrow\dfrac{x+4}{6}>0\Leftrightarrow x+4>0\Leftrightarrow x>-4\)
\(\forall x\ne\pm2;x\ne-3\)
