Bài I:
1: Thay x=-3 vào A, ta được:
\(A=\dfrac{-3+1}{\left(-3\right)^2-2\cdot\left(-3\right)}=\dfrac{-2}{9+6}=\dfrac{-2}{15}\)
2: \(B=\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}-\dfrac{16}{4-x^2}\)
\(=\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}+\dfrac{16}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+4-x^2+4x-4+16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x+16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x-2}\)
3: \(P=B:A=\dfrac{8}{x-2}:\dfrac{x+1}{x\left(x-2\right)}\)
\(=\dfrac{8}{x-2}\cdot\dfrac{x\left(x-2\right)}{x+1}=\dfrac{8x}{x+1}\)
Để P là số tự nhiên thì \(\left\{{}\begin{matrix}P>=0\\8x⋮x+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{8x}{x+1}>=0\\8x+8-8⋮x+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{x+1}>=0\\x+1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>0\\x< -1\end{matrix}\right.\\x\in\left\{0;-2;1;-3;3;-5;7;-9\right\}\end{matrix}\right.\)
=>\(x\in\left\{-2;-3;1;3;-5;7;-9\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{-3;1;3;-5;7;-9\right\}\)
Bài II:
1: \(7\left(3x-2\right)+4\left(5-3x\right)=15\)
=>\(21x-14+20-12x=15\)
=>9x+6=15
=>9x=9
=>x=1
2: \(x-\dfrac{x-1}{3}=\dfrac{2x+1}{5}\)
=>\(\dfrac{15x-5\left(x-1\right)}{15}=\dfrac{3\left(2x+1\right)}{15}\)
=>15x-5x+5=6x+3
=>6x+3=10x+5
=>-4x=2
=>\(x=-\dfrac{1}{2}\)
3: \(12-3\left(x-2\right)^2=\left(x+2\right)\left(1-3x\right)\)
=>\(12-3\left(x^2-4x+4\right)=x-3x^2+2-6x\)
=>\(-3x^2-5x+2=12-3x^2+12x-12\)
=>-5x+2=12x
=>-17x=-2
=>\(x=\dfrac{2}{17}\)
