Question

[Ju Moon Adn]

rút gọn phân thức

1.\(\dfrac{3\left|x-4\right|}{3x^2-3x-36}\)

2.\(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)

3.\(\dfrac{x^2+5x+6}{x^2+4x+4}\)

Answer 1

[Nguyễn Huy Tú]

a, \(\dfrac{3\left|x-4\right|}{3x^2-3x-36}=\dfrac{3\left|x-4\right|}{3\left(x^2-x-12\right)}=\dfrac{\left|x-4\right|}{x^2-x-12}\)

b, \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}=\dfrac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}\)

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

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

c, \(\dfrac{x^2+5x+6}{x^2+4x+4}=1+\dfrac{x+2}{x^2+4x+4}=1+\dfrac{x+2}{\left(x+2\right)^2}\)

\(=1+\dfrac{1}{x+2}=\dfrac{x+3}{x+2}\)

Answer 2

[Nguyễn Huy Tú]

a, tiếp:

+) Xét \(x\ge4\) có:

\(\dfrac{x-4}{x^2-x-12}=\dfrac{x-4}{x^2-4x+3x-12}=\dfrac{x-4}{x\left(x-4\right)+3\left(x-4\right)}\)

\(=\dfrac{x-4}{\left(x+3\right)\left(x-4\right)}=\dfrac{1}{x+3}\)

+) Xét x < 4 có:
\(\dfrac{4-x}{x^2-x-12}=\dfrac{4-x}{x^2-4x+3x-12}=\dfrac{4-x}{x\left(x-4\right)+3\left(x-4\right)}\)

\(=\dfrac{4-x}{\left(x+3\right)\left(x-4\right)}=\dfrac{-\left(x-4\right)}{\left(x+3\right)\left(x-4\right)}=\dfrac{-1}{x+3}\)