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a: Sửa đề: 6x^3
\(\dfrac{6x^3-4x^2+3x-2}{3x-2}=\dfrac{2x^2\left(3x-2\right)+3x-2}{3x-2}=2x^2+1\)
b: \(\dfrac{6x^3-3x^2+4x+2}{3x^2+2}\)
\(=\dfrac{6x^3+4x-3x^2-2+4}{3x^2+2}\)
\(=2x-1+\dfrac{4}{3x^2+2}\)
ĐKXĐ: \(x\ne\left\{-\dfrac{1}{3};\dfrac{1}{3};0;-\dfrac{4}{3}\right\}\)
\(M=\left(\dfrac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right):\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\left(\dfrac{x\left(3x+5\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{1-3x}{2\left(1+3x\right)}\)
\(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)\(\left(ĐKXĐ:x\ne\pm\dfrac{1}{3}\right)\)
\(=\left[\dfrac{-3x\left(3x+1\right)+2x\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}\right]:\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)
\(=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{-3x^2-5x}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{-x\left(3x+5\right).\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right).2x\left(3x+5\right)}\)
\(=\dfrac{1-3x}{2\left(3x+1\right)}\)
\(=\dfrac{1-3x}{6x+2}\)
3x2 - 6x = 3x(x-2)