*) Xét \(\frac{x^2-2x-3}{x^2+x}\) và \(\frac{x-3}{x}\)
Ta có: \(\left(x^2-2x-3\right).x=x^3-2x^2-3x\)
\(\left(x^2+x\right).\left(x-3\right)=x^3-3x^2+x^2-3x=x^3-2x^2-3x\)
\(\Rightarrow\left(x^2-2x-3\right).x=\left(x^2+x\right)\left(x-3\right)\)
\(\Rightarrow\frac{x^2-2x-3}{x^2+x}=\frac{x-3}{x}\) (1)
*) Xét \(\frac{x-3}{x}\) và \(\frac{x^2-4x+3}{x^2-x}\)
Ta có:
\(\left(x-3\right).\left(x^2-x\right)=x^3-x^2-3x^2+3x=x^3-4x^2+3x\)
\(x.\left(x^2-4x+3\right)=x^3-4x^2+3x\)
\(\Rightarrow\left(x-3\right).\left(x^2-x\right)=x.\left(x^2-4x+3\right)\)
\(\Rightarrow\) \(\frac{x-3}{x}\) = \(\frac{x^2-4x+3}{x^2-x}\) (2)
Từ (1) và (2) \(\Rightarrow\)\(\frac{x^2-2x-3}{x^2+x}\) = \(\frac{x-3}{x}\) = \(\frac{x^2-4x+3}{x^2-x}\)