Ta có: \(\dfrac{20}{x}-\dfrac{20}{x+2}=\dfrac{10}{3}\)
Suy ra: \(10x\left(x+2\right)=60\left(x+2\right)-60x\)
\(\Leftrightarrow10x^2+20x-120=0\)
\(\Leftrightarrow x^2+2x-12=0\)
\(\Leftrightarrow\left(x+1\right)^2=13\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{13}-1\left(nhận\right)\\x=-\sqrt{13}-1\left(nhận\right)\end{matrix}\right.\)