Question

Qui đồng mẫu thức các phân thức:

\(\dfrac{-3}{x^2+6x+8};\dfrac{5}{x^2-16}\) và \(\dfrac{1}{x^2-2x-8}\)

Answer 1

MTC=(x+2)(x+4)(x-2)(x-4)

\(\dfrac{-3}{x^2+6x+8}=\dfrac{-3\left(x-2\right)\left(x-4\right)}{\left(x+2\right)\left(x-2\right)\left(x+4\right)\left(x-4\right)}\)

\(\dfrac{5}{x^2-16}=\dfrac{5\left(x+2\right)\left(x-2\right)}{\left(x-4\right)\left(x+4\right)\left(x-2\right)\left(x+2\right)}\)

\(\dfrac{1}{x^2-2x-8}=\dfrac{1}{\left(x-4\right)\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(x+4\right)\left(x-4\right)}\)