\(a,=\dfrac{x^2+4x+4}{6\left(x+2\right)}=\dfrac{\left(x+2\right)^2}{6\left(x+2\right)}=\dfrac{x+2}{6}\\ b,=\dfrac{3x-2-x+4-2x}{x-1}=\dfrac{2}{x-1}\\ c,=\dfrac{x^2-2x+2}{\left(x-2\right)\left(x+2\right)}\\ d,=\dfrac{x^2+2+3x-3-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2}{x^2+x+1}\)
\(a,\dfrac{x^2}{6x+12}+\dfrac{4x+4}{6x+12}=\dfrac{x^2+4x+4}{6x+12}=\dfrac{\left(x+2\right)^2}{6\left(x+2\right)}=\dfrac{x+2}{6}\\ b,\dfrac{3x-2}{x-1}+\dfrac{x}{1-x}+\dfrac{4-2x}{x-1}=\dfrac{3x-2}{x-1}-\dfrac{x}{x-1}+\dfrac{4-2x}{x-1}=\dfrac{3x-2-x+4-2x}{x-1}=\dfrac{2}{x-1}\)
\(c,\dfrac{x}{x+2}+\dfrac{2}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2-2x+2}{\left(x+2\right)\left(x-2\right)}\)
\(d,\dfrac{x^2+2}{x^3-1}+\dfrac{3}{x^2+x+1}+\dfrac{1}{1-x}=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1}{x-1}=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+2+3x-3-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2}{x^2+x+1}\)
