a, \(\dfrac{x+3}{3x-3}\) + \(\dfrac{2-x}{4x-4}\) 

đkxđ: \(\left\{{}\begin{matrix}3x-3\ne0\\4x-4\ne0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}3.\left(x-1\right)\ne0\\4.\left(x-1\right)\ne0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-1\ne0\\x-1\ne0\end{matrix}\right.\) ⇒ \(x\) ≠1

A = \(\dfrac{x+3}{3x-3}\) + \(\dfrac{2-x}{4x-4}\)

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

A = \(\dfrac{4.\left(x+3\right)}{4.3.\left(x-1\right)}\) + \(\dfrac{3.\left(2-x\right)}{3.4.\left(x-1\right)}\)

A = \(\dfrac{4x+12}{12\left(x-1\right)}\) + \(\dfrac{6-3x}{12\left(x-1\right)}\)

A = \(\dfrac{4x+12+6-3x}{12\left(x-1\right)}\)

A = \(\dfrac{\left(4x-3x\right)+\left(12+6\right)}{12\left(x-1\right)}\)

A = \(\dfrac{x+18}{12.\left(x-1\right)}\)

b, \(\dfrac{3}{x+2}-\dfrac{6}{x-1}\) 

Đkxđ: \(\left\{{}\begin{matrix}x+2\ne0\\x-1\ne0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x\ne-2\\x\ne1\end{matrix}\right.\)

B = \(\dfrac{3}{x+2}\) - \(\dfrac{6}{x-1}\)

B = \(\dfrac{3.\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}\) - \(\dfrac{6.\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

B = \(\dfrac{3x-3}{\left(x+2\right)\left(x-1\right)}\) - \(\dfrac{6x+12}{\left(x-1\right)\left(x+2\right)}\)

B = \(\dfrac{3x-3-\left(6x+12\right)}{\left(x+2\right)\left(x-1\right)}\)

B = \(\dfrac{3x-3-6x-12}{\left(x+2\right)\left(x-1\right)}\)

B = \(\dfrac{\left(3x-6x\right)-\left(12+3\right)}{\left(x+2\right)\left(x-1\right)}\)

B = \(\dfrac{-3x-15}{\left(x+2\right)\left(x-1\right)}\)

c, C = \(\dfrac{1}{3x-1}\) - \(\dfrac{1}{3x+1}\) - \(\dfrac{2x-3}{1-9x^2}\)

Đkxđ: \(\left\{{}\begin{matrix}3x-1\ne0\\3x+1\ne0\\1-9x^2\ne0\end{matrix}\right.\)  ⇒ \(\left\{{}\begin{matrix}3x\ne1\\3x\ne-1\\\left(1-3x\right)\left(1+3x\right)\ne0\end{matrix}\right.\)⇒ \(\left\{{}\begin{matrix}x\ne\dfrac{1}{3}\\x\ne-\dfrac{1}{3}\end{matrix}\right.\)

C = \(\dfrac{1}{3x-1}\) - \(\dfrac{1}{3x+1}\) + \(\dfrac{2x-3}{\left(3x-1\right)\left(1+3x\right)}\)

C = \(\dfrac{\left(3x+1\right)}{\left(3x+1\right)\left(3x-1\right)}\) - \(\dfrac{3x-1}{\left(3x+1\right)\left(3x-1\right)}\)+\(\dfrac{2x-3}{\left(3x-1\right)\left(3x+1\right)}\)

C = \(\dfrac{3x+1-\left(3x-1\right)+2x-3}{\left(3x+1\right)\left(3x-1\right)}\)

C = \(\dfrac{3x+1-3x+1+2x-3}{\left(3x+1\right)\left(3x-1\right)}\)

C = \(\dfrac{\left(3x-3x+2x\right)+\left(1+1-3\right)}{\left(3x+1\right)\left(3x-1\right)}\)

C = \(\dfrac{2x-1}{\left(3x+1\right)\left(3x-1\right)}\)

d, D = \(\dfrac{6x}{x^3+1}\) - \(\dfrac{1}{x^2-x+1}\) + \(\dfrac{2}{x+1}\)

    đkxđ: \(x^3\) + 1 ≠ 0; ⇒ \(x\) ≠ -1; \(x\) + 1 ≠ 0 ⇒ \(x\) ≠ -1;

        \(x^2\) - \(x\) + 1 = (\(x^2\) - 2.\(\dfrac{1}{2}\)\(x\) + \(\dfrac{1}{4}\)) + \(\dfrac{3}{4}\) = (\(x\) - \(\dfrac{1}{2}\))² + \(\dfrac{3}{4}\) > 0 ∀\(x\)

      vậy \(x\) ≠ - 1

D = \(\dfrac{6x}{x^3+1}\) - \(\dfrac{1}{x^2-x+1}\) + \(\dfrac{2}{x+1}\)

D = \(\dfrac{6x}{\left(x+1\right).\left(x^2-x+1\right)}\) - \(\dfrac{1}{x^2-x+1}\) + \(\dfrac{2}{x+1}\)

D = \(\dfrac{6x}{\left(x+1\right)\left(x^2-x+1\right)}\) - \(\dfrac{x+1}{\left(x+1\right).\left(x^2-x+1\right)}\)+\(\dfrac{2.\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

D = \(\dfrac{6x-\left(x+1\right)+2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

D = \(\dfrac{6x-x-1+2x^2-2x+2}{\left(x+1\right)\left(x^2-x+1\right)}\)

D = \(\dfrac{2x^2+\left(6x-x-2x\right)+\left(2-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

D = \(\dfrac{2x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

D = \(\dfrac{2x^2+2x+x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

D = \(\dfrac{2.\left(x+1\right)x+\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\)

D = \(\dfrac{\left(x+1\right)\left(2x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\)

D = \(\dfrac{2x+1}{x^2-x+1}\)