Question

Ba phân thức sau có bằng nhau không?
\(\dfrac{x^2-2x-3}{x^2+x};\dfrac{x-3}{x};\dfrac{x^2-4x+3}{x^2-x}\)

Answer 1

\(\left\{{}\begin{matrix}\dfrac{x^2-2x-3}{x^2+x}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\\\dfrac{x-3}{x}\\\dfrac{x^2-4x+3}{x^2-x}=\dfrac{\left(x-3\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\end{matrix}\right.\)

Vậy \(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)

Answer 2

\(ĐK:x\ne0;x\ne\pm1\\ \dfrac{x^2-2x-3}{x^2+x}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\\ \dfrac{x^2-4x+3}{x^2-x}=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\)

Do đó 3 phân thức trên bằng nhau