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