`3) A=(\frac{x^3}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{2+x}):(x+2+\frac{10-x^2}{x-2})`
ĐKXĐ: `x\ne0, ±2.`
` A=(\frac{x^3}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{2+x}):(x+2+\frac{10-x^2}{x-2})`
`A=(\frac{x.x^2}{x(x^2-4)}+\frac{3.2}{3.(2-x)}+\frac{1}{2+x}):(\frac{x^2-4}{x-2}+\frac{10-x^2}{x-2})`
`A=(\frac{x^2}{(x-2)(x+2)}-\frac{2(x+2)}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}):(\frac{x^2-4+10-x^2}{x-2})`
`A=(\frac{x^2-2x-4+x-2}{(x-2)(x+2)}):(\frac{6}{x-2})`
`A=\frac{x^2−x−6}{(x-2)(x+2)}.\frac{x-2}{6}`
`A=\frac{x^2−x−6}{6(x+2)}`
`A=\frac{x^2+2x-3x−6}{6(x+2)}`
`A=\frac{x(x+2)-3(x+2)}{6(x+2)}`
`A=\frac{(x+2)(x-3)}{6(x+2)}`
`A=\frac{x-3}{6}.`
Vậy `A=\frac{x-3}{6}.`
