Question

Answer 1

a: \(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{-5x-5}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}=\dfrac{-5}{x-3}\)

b: \(x^2-x-2=0\)

=>(x-2)(x+1)=0

=>x=2(nhận) hoặc x=-1(loại)

Thay x=2 vào A, ta được: 

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

c: Để A=1/2 thì \(\dfrac{-5}{x-3}=\dfrac{1}{2}\)

=>x-3=-10

hay x=-7