Question

\(\frac{x}{x-2}\)+\(\frac{x+2}{x}\)>2

Answer 1

\(\frac{x}{x-2}+\frac{x+2}{x}\)

=\(\frac{x^2}{x\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}\)

=\(\frac{2x^2-4}{x\left(x-2\right)}\)

>\(\frac{2x^2-8}{x\left(x-2\right)}\)

=\(\frac{2\left(x^2-4\right)}{x\left(x-2\right)}\)

=\(\frac{2\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}=\frac{2\left(x+2\right)}{x}=\frac{2x+4}{x}\)

=\(2+\frac{4}{x}>2\)

⇒\(\frac{x}{x-2}+\frac{x+2}{x}>2\)