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a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2}{x^2-4}\)
\(ĐKXĐ:x\ne\pm2\)
\(a,A=\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x+2+x-2+x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x+x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(2+x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x}{x-2}\)
\(b,A=\frac{x}{x-2}\)
\(=\frac{x-2+2}{x-2}\)
\(=\frac{x-2}{x-2}+\frac{2}{x-2}\)
\(=1+\frac{2}{x-2}\)
\(\text{Để A có giá trị nguyên thì:2⋮ x-2}\)
\(\text{hay }x-2\inƯ\left(2\right)=\left\{-1;1;-2;2\right\}\)
\(\Rightarrow x\in\left\{1;3;0;4\right\}\left(tm\right)\)
\(\text{Vậy }x\in\left\{1;3;0;4\right\}\) \(\text{thì A có giá trị nguyên.}\)
\(A=\left[\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right]:\left[x-2+\frac{10-x^2}{x+2}\right]\) ĐKXĐ : \(x\ne0;x\ne\pm2\)
\(A=\left[\frac{x^2}{x\left(x+2\right)\left(x-2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{x^2-4}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\left[\frac{3x^2}{3x\left(x+2\right)\left(x-2\right)}-\frac{6x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}+\frac{3x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}\right]:\frac{6}{x+2}\)
\(A=\left[\frac{3x^2-6x^2-12x+3x^2+6x}{3x\left(x+2\right)\left(x-2\right)}\right].\frac{x+2}{6}\)
\(A=\frac{-x}{3x\left(x-2\right)}\)
\(A=\frac{-1}{3x-6}\)
\(A=\left(\frac{x^2-16}{x-4}+1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)
\(=\left(x+5\right):\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x^2+x-2x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x+3}{x+1}\right)=\frac{x+3}{\left(x+5\right)\left(x+1\right)}\)