Question

  Giúp mik voii

Answer 1

a: ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)

\(A=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{x^2-9}:\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)

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

b: \(x^2-x-2=0\)

=>\(\left(x-2\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

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

c: A=1/2

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

=>x-3=-10

=>x=-7(nhận)