Ta có △ABC vuông tại A đường cao AH⇒AB.AC=AH.BC=\(\dfrac{24}{5}.10=48\Rightarrow AB=\dfrac{48}{AC}\Rightarrow AB^2=\dfrac{2304}{AC^2}\)
Ta có △ABC vuông tại A⇒BC2=AB2+AC2⇒\(100=\dfrac{2304}{AC^2}+AC^2=\dfrac{AC^4+2304}{AC^2}\Rightarrow AC^4+2304=100AC^2\Rightarrow AC^4-100AC^2+2304=0\Rightarrow\left[{}\begin{matrix}AC=8\left(cm\right)\\AC=6\left(cm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}AB=\dfrac{48}{AC}=\dfrac{48}{8}=6\left(cm\right)\\AB=\dfrac{48}{AC}=\dfrac{48}{6}=8\left(cm\right)\end{matrix}\right.\)Vậy (AB,AC)=(6;8);(8;6)