a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

dk 3x+2 

P= \(\frac{x\left(3x-1\right)}{3x+2}.\frac{3x+2}{\left(3x-1\right)x^2+4\left(3x-1\right)}=\frac{x\left(3x-1\right)}{3x+2}.\frac{3x+2}{\left(3x-1\right)\left(x^2+4\right)}=\)\(\frac{x}{x^2+4}\)

dk \(\hept{\begin{cases}3x-1\ne0\\3x+2\ne0\end{cases}< =>\hept{\begin{cases}x\ne\frac{1}{3}\\x\ne\frac{-2}{3}\end{cases}}}\)(1)

P(x²+4) = x <=> Px²-x+4P=0

để phương trình trên có nghiệm thỏa mãn (1) <=> \(\hept{\begin{cases}P\frac{1}{3^2}-\frac{1}{3}+4P\ne0\\P\frac{4}{9}+\frac{2}{3}+4P\ne0\\1^2-4.P.\left(4P\right)\ge0\end{cases}< =>\hept{\begin{cases}P\ne\frac{3}{37}\\P\ne\frac{-3}{20}\\\frac{-1}{4}\le P\le\frac{1}{4}\end{cases}}}\)

Vậy P max = 1/4 khi \(\frac{1}{4}x^2-x+1=0< =>x=2\)

P min = -1/4 khi \(\frac{-1}{4}x^2-x-1=0< =>x=-2\)

ĐKXĐ: \(x\ne\pm3\)

\(P=\left[\dfrac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)

\(=\left[\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)

\(=\dfrac{x+3}{x^2+9}:\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}=\dfrac{x+3}{x^2+9}.\dfrac{\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)

Ý 2 mình k hiểu ý bạn lắm

\(P=\dfrac{x+3}{x-3}=\dfrac{x-3+6}{x-3}=1+\dfrac{6}{x-3}\in Z\)

\(\Leftrightarrow\left(x-3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Kết hợp vs ĐKXĐ \(\Rightarrow x\in\left\{0;1;2;4;5;6;9\right\}\)

\(A=\left[\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right]:\left[x-2+\frac{10-x^2}{x+2}\right]\) ĐKXĐ : \(x\ne0;x\ne\pm2\)

\(A=\left[\frac{x^2}{x\left(x+2\right)\left(x-2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{x^2-4}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(A=\left[\frac{3x^2}{3x\left(x+2\right)\left(x-2\right)}-\frac{6x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}+\frac{3x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}\right]:\frac{6}{x+2}\)

\(A=\left[\frac{3x^2-6x^2-12x+3x^2+6x}{3x\left(x+2\right)\left(x-2\right)}\right].\frac{x+2}{6}\)

\(A=\frac{-x}{3x\left(x-2\right)}\)

\(A=\frac{-1}{3x-6}\)

Câu 4: 

\(=\dfrac{a\left(a-b\right)-c\left(a-b\right)}{a\left(a+b\right)-c\left(a+b\right)}=\dfrac{a-b}{a+b}\)