\(\dfrac{x}{x-2}\) =\(\dfrac{x+2}{x}\) > 2
MTC : x(x - 2)
\(\dfrac{x^2}{x\left(x-2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}=\dfrac{2x\left(x-2\right)}{x\left(x-2\right)}\)
\(\Leftrightarrow\) x2 + (x - 2) (x + 2) > 2x(x - 2)
\(\Leftrightarrow\) x2 + x2 - 4 > 2x2 - 4x
\(\Leftrightarrow\) x2 + x2 + 4 > 2x2 - 4x
\(\Leftrightarrow\) x2 + x2 - 2x2 + 4x > -4
\(\Leftrightarrow\) 4x > -4
\(\Leftrightarrow\) x > \(\dfrac{-4}{4}=-1\)
Vay x > - 1
\(\dfrac{x}{x-2}+\dfrac{x+2}{x}>2\) \(\left(ĐK:x\ne0;x\ne2\right)\)
⇔ \(\dfrac{x^2}{x\left(x-2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}>2\)
⇔ \(\dfrac{x^2+x^2-4}{x\left(x-2\right)}>2\)
⇔ \(\dfrac{2x^2-4}{x^2-2x}-2>0\)
⇔ \(2x^2-4-2x^2+4x>0\)
\(\Leftrightarrow4\left(x-1\right)>0\)
⇔ \(x-1>0\)
⇔ \(x>1\)
Kết hợp ĐK, ta có \(x>1\) và \(x\ne2\)
