a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)
b: \(P=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-4-x-8}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)
c: \(P=-\dfrac{3}{4}\)
=>\(\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)
=>4(x-4)=-3(x-2)
=>4x-16=-3x+6
=>7x=22
=>\(x=\dfrac{22}{7}\left(nhận\right)\)
d: Để P nguyên thì \(x-4⋮x-2\)
=>\(x-2-2⋮x-2\)
=>\(-2⋮x-2\)
=>\(x-2\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{3;1;4;0\right\}\)
e: \(x^2-9=0\)
=>\(x^2=9\)
=>\(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Thay x=3 vào P, ta được:
\(P=\dfrac{3-4}{3-2}=\dfrac{-1}{1}=-1\)
