Ta có: \(\dfrac{3x+2}{x^2-2x+1}-\dfrac{6}{x^2-1}-\dfrac{3x-2}{x^2+2x+1}\)
\(=\dfrac{3x+2}{\left(x-1\right)^2}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}-\dfrac{3x-2}{\left(x+1\right)^2}\)
\(=\dfrac{\left(3x+2\right)\left(x^2+2x+1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}-\dfrac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}-\dfrac{\left(3x-2\right)\left(x^2-2x+1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}\)
\(=\dfrac{3x^3+6x^2+3x+2x^2+4x+2-6\left(x^2-1\right)-\left(3x^3-6x^2+3x-2x^2+4x-2\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}\)
\(=\dfrac{3x^3+8x^2+7x+2-6x^2+6-\left(3x^3-8x^2+7x-2\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}\)
\(=\dfrac{3x^3+2x^2+7x+8-3x^3+8x^2-7x+2}{\left(x-1\right)^2\cdot\left(x+1\right)^2}\)
\(=\dfrac{10x^2+10}{\left(x-1\right)^2\cdot\left(x+1\right)^2}\)
