Ta có \(\left\{{}\begin{matrix}S_{ABC}=1\\S_{ABC}=\dfrac{AH.BC}{2}\end{matrix}\right.\Rightarrow\dfrac{AH.BC}{2}=1\Rightarrow\dfrac{2\sqrt{5}}{5}.BC=2\Rightarrow BC=\sqrt{5}}\)Ta có △ABC vuông tại A đường cao AH⇒AB.AC=AH.BC=\(\dfrac{2\sqrt{5}}{5}.\sqrt{5}=2\Rightarrow AB=\dfrac{2}{AC}\Rightarrow AB^2=\dfrac{4}{AC^2}\)
Ta có △ABC vuông tại A⇒BC2=AB2+AC2⇒\(5=\dfrac{4}{AC^2}+AC^2=\dfrac{4+AC^4}{AC^2}\Rightarrow AC^4+4=5AC^2\Rightarrow AC^4-5AC^2+4=0\Rightarrow\left[{}\begin{matrix}AC=1\\AC=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}AB=\dfrac{2}{AC}=\dfrac{2}{1}=2\\AB=\dfrac{2}{AC}=\dfrac{2}{2}=1\end{matrix}\right.\)Vậy (AB,AC)=(2;1);(1;2)
