Question

Answer 1

Sorry bạn phần c và d mk chưa nghĩ ra cách

Answer 2

a: Ta có: \(P=\left(\dfrac{2}{x+3}+\dfrac{1}{x-3}+\dfrac{6x}{9-x^2}\right):\left(\dfrac{2x-2}{x-3}-1\right)\)

\(=\dfrac{2x-6+x+3-6x}{\left(x+3\right)\left(x-3\right)}:\dfrac{2x-2-x+3}{x-3}\)

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

\(=\dfrac{-3}{x+3}\)

Để P<0 thì x+3<0

hay x<-3

b: Để \(P=-\dfrac{1}{3}\) thì x+3=9

hay x=6