\(\text{Ta có : }\dfrac{x^2-2x-3}{x^2+x}\\ =\dfrac{x^2+x-3x-3}{x\left(x+1\right)}\\ =\dfrac{\left(x^2+x\right)-\left(3x+3\right)}{x\left(x+1\right)}\\ \\ =\dfrac{x\left(x+1\right)-3\left(x+1\right)}{x\left(x+1\right)}\\ \\ =\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}\\ \\ =\dfrac{x-3}{x}\text{ }\text{ }\text{ }\left(1\right)\)
\(\dfrac{x^2-4x+3}{x^2-x}\\ =\dfrac{x^2-x-3x+3}{x\left(x-1\right)}\\ \\ =\dfrac{\left(x^2-x\right)-\left(3x-3\right)}{x\left(x-1\right)}\\ \\ =\dfrac{x\left(x-1\right)-3\left(x-1\right)}{x\left(x-1\right)}\\ \\ =\dfrac{\left(x-3\right)\left(x-1\right)}{x\left(x-1\right)}\\ \\ =\dfrac{x-3}{x}\text{ }\text{ }\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)
Vậy 3 phân thức \(\dfrac{x^2-2x-3}{x^2+x};\dfrac{x-3}{x};\dfrac{x^2-4x+3}{x^2-x}\) bằng nhau
Giả sử :
\(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)
\(\Leftrightarrow\) \(\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{x-3}{x}=\dfrac{x-3}{x}=\dfrac{x-3}{x}\)
Vậy 3 thức trên bằng nhau
