Question

Giúp mình bài 2 với ạ

Answer 1

​a: Khi x=3 thì \(A=\dfrac{3+3}{9-4}=\dfrac{6}{5}\)

b: \(B=\dfrac{x^2-x-2-x^2+2x}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}\)

c: \(P=\dfrac{B}{A}=\dfrac{1}{x+2}:\dfrac{x+3}{x^2-4}=\dfrac{x-2}{x+3}\)

Để P<1 thì P-1<0

=>(x-2-x-3)/(x+3)<0

=>x+3>0

=>x>-3