Question

Answer 1

a: \(B=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

=x-1

b: Để B=0 thì x-1=0

hay x=1

Answer 2

\(ĐK:x\ne\pm2\\ a,B=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\\ B=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x^2-4\right)\left(x-1\right)}{x^2-4}=x-1\\ b,B=0\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\\ c,B>0\Leftrightarrow x-1>0\Leftrightarrow x>1;x\ne2\)