\(A=\dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4}\)
\(A=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(A=\dfrac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(A=\dfrac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)
\(A=\dfrac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
Ta có: -2 < x < 2
=> x thuộc { -1 ; 0 ; 1 }
Mà x khác -1 nên x = 0 ; x = 1
Với x = 0 thì \(A=\dfrac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(0+1\right)^2}{\left(0-2\right)\left(0+2\right)}=\dfrac{1}{-4}\)
=> A có giá trị âm
Với x = 1 thì \(A=\dfrac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(1+1\right)^2}{\left(1-2\right)\left(1+2\right)}=\dfrac{4}{-3}\)
=> A có giá trị âm
Vậy với -2 < x < 2 ; x khác -1 thì A có giá trị âm
