a: |x-2|=1
=>\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=3 vào A, ta được:
\(A=\dfrac{3+1}{3^2-2\cdot3}=\dfrac{4}{9-6}=\dfrac{4}{3}\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1}{1^2-2\cdot1}=\dfrac{2}{-1}=-2\)
b: \(B=\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}-\dfrac{16}{4-x^2}\)
\(=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{16}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+4-x^2+4x-4+16}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{8x+16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x-2}\)
c: \(P=B:A=\dfrac{8}{x-2}:\dfrac{x+1}{x\left(x-2\right)}\)
\(=\dfrac{8}{x-2}\cdot\dfrac{x\left(x-2\right)}{x+1}=\dfrac{8x}{x+1}\)
d: Để P là số tự nhiên thì \(\left\{{}\begin{matrix}8x⋮x+1\\\dfrac{x}{x+1}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x+8-8⋮x+1\\\left[{}\begin{matrix}x>=0\\x< -1\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\\\left[{}\begin{matrix}x>=0\\x< -1\end{matrix}\right.\end{matrix}\right.\)
=>\(x\in\left\{0;-2;1;-3;3;-5;7;-9\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{1;-3;3;-5;7;-9\right\}\)
