Question

Answer 1

a: \(A=\left(\dfrac{x}{x+3}+\dfrac{-\left(x-3\right)}{x+3}\cdot\dfrac{x^2+3x+9}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{3}{x+3}\)

\(=\left(\dfrac{x}{x+3}-\dfrac{x^2+3x+9}{\left(x+3\right)^2}\right)\cdot\dfrac{x+3}{3}\)

\(=\dfrac{x^2+3x-x^2-3x-9}{\left(x+3\right)^2}\cdot\dfrac{x+3}{3}\)

\(=\dfrac{-9}{3\left(x+3\right)}=\dfrac{-3}{x+3}\)

b: Thay x=-1/2 vào A, ta được:

\(A=\dfrac{-3}{3-\dfrac{1}{2}}=-3:\dfrac{5}{2}=-\dfrac{6}{5}\)

c: Để A<0 thì x+3>0

hay x>-3

Kết hợp ĐKXĐ, ta được: x>-3 và x<>3