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Lời giải:
$\frac{x^3-3x^2-x+3}{x^2-3x}=\frac{x^2(x-3)-(x-3)}{x(x-3)}=\frac{(x-3)(x^2-1)}{x(x-3)}=\frac{x^2-1}{x}$
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Ta có: \(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}\)
= \(\frac{\left(y-x\right)\left(y+x\right)}{\left(x-y\right)^3}\)
=\(-\frac{x+y}{\left(x-y\right)^2}\)
=\(-\frac{x+y}{x^2-2xy+y^2}\)
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a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
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\(\dfrac{x^3+3x^2-2}{x^3+3x+4}\)
\(=\dfrac{x^3+x^2+2x^2+2x-2x-2}{x^3-x+4x+4}\)
\(=\dfrac{\left(x+1\right)\left(x^2+2x-2\right)}{\left(x+1\right)\left(x^2-x+4\right)}\)
\(=\dfrac{x^2+2x-2}{x^2-x+4}\)
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\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2\left(x+1\right)+\left(x+1\right)}{3\left(x^2+2x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left(x^2+x+x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left[x\left(x+1\right)+\left(x+1\right)\right]}\)
\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left(x+1\right)\left(x+1\right)}=\frac{x^2+1}{3\left(x+1\right)}\)
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a)\(\frac{x^2+3x+2}{3x+6}=\frac{x^2+2x+x+2}{3\cdot\left(x+2\right)}=\frac{\left(x^2+2x\right)+\left(x+2\right)}{3\cdot\left(x+2\right)}=\frac{x\cdot\left(x+2\right)+\left(x+2\right)}{3\cdot\left(x+2\right)}\)
\(=\frac{\left(x+2\right)\cdot\left(x+1\right)}{3\cdot\left(x+2\right)}=\frac{x+1}{3}\)
b) \(\frac{2x^2+x-1}{6x-3}=\frac{2x^2+2x-x-1}{3\cdot\left(2x-1\right)}=\frac{\left(2x^2+2x\right)-\left(x+1\right)}{3\cdot\left(2x-1\right)}\)
\(=\frac{2x\cdot\left(x+1\right)-\left(x+1\right)}{3\cdot\left(2x-1\right)}=\frac{\left(2x-1\right)\cdot\left(x+1\right)}{3\cdot\left(2x-1\right)}=\frac{x+1}{3}\)
\(\frac{x^3+125}{x^2-3x-40}=\frac{x^3+5^3}{\left(x^2+5x\right)-\left(8x+40\right)}=\frac{\left(x+5\right)\left(x^2-5x+25\right)}{x\left(x+5\right)-8\left(x+5\right)}\)
\(=\frac{\left(x+5\right)\left(x^2-5x+25\right)}{\left(x+5\right)\left(x-8\right)}=\frac{x^2-5x+25}{x-8}\)