Đáp án:

$H.3 \left\{\begin{array}{l} x=\dfrac{25}{4}\\ y=\dfrac{5\sqrt{41}}{4}\end{array} \right.\\ H.4 \left\{\begin{array}{l} x=\dfrac{7}{2}\\ y=56 \end{array} \right.\\ H.5 \left\{\begin{array}{l} x=\dfrac{63\sqrt{130}}{130}\\ y=\sqrt{130}\end{array} \right.\\ H.6 \left\{\begin{array}{l} y=10\\ x=7,5\end{array} \right.$

Giải thích các bước giải:

$H.3$

$\Delta ABC$ vuông tại $A$, đường cao $AH$

$\Rightarrow \left\{\begin{array}{l} AH^2=BH.CH\\ AC^2=AH^2+HC^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 4x=5^2\\ y^2=5^2+x^2\end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{25}{4}\\ y=\dfrac{5\sqrt{41}}{4}\end{array} \right.\\ H.4$

$\Delta ABC$ vuông tại $A$, đường cao $AH$

$\Rightarrow \dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\\ \Leftrightarrow \dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{(4AB)^2}\\ \Leftrightarrow \dfrac{1}{14^2}=\dfrac{17}{16AB^2}\\ \Leftrightarrow AB=\dfrac{7\sqrt{17}}{2}\\ \Rightarrow AC=4AB=14\sqrt{17}$

$\Delta ABC$ vuông tại $A$

$\Rightarrow x=BH=\sqrt{AB^2-AH^2}=\dfrac{7}{2}\\ \Rightarrow y=CH=\sqrt{AC^2-AH^2}=56\\ H.5$

$\Delta ABC$ vuông tại $A$, đường cao $AH$

$\Rightarrow \left\{\begin{array}{l} \dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\\ BC^2=AB^2+AC^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \dfrac{1}{x^2}=\dfrac{1}{7^2}+\dfrac{1}{9^2}\\ BC^2=7^2+9^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{63\sqrt{130}}{130}\\ y=\sqrt{130}\end{array} \right.\\ H.6$

$\Delta ABC$ vuông tại $A$, đường cao $AH$

$\Rightarrow \left\{\begin{array}{l} AH^2=BH.CH\\ AC^2=AH^2+CH^2\\ AB^2=AH^2+BH^2\end{array} \right.\\ \Leftrightarrow \Rightarrow \left\{\begin{array}{l} 6^2=4,5.CH\\ y^2=6^2+CH^2\\ x^2=6^2+4,5^2\end{array} \right.\\ \Leftrightarrow \Rightarrow \left\{\begin{array}{l} CH=8\\ y=10\\ x=7,5\end{array} \right.$