Đáp án:
$\begin{array}{l}
B1)a)\\
M{H^2} = PH.NH\\
\Rightarrow {12^2} = 16.x\\
\Rightarrow x = 9\\
\Rightarrow y = \sqrt {{{12}^2} + {x^2}} = \sqrt {{{12}^2} + {9^2}} = 15\\
\text{Vậy}\,x = 9;y = 15\\
b)\\
AB = \sqrt {B{C^2} - A{C^2}} = \sqrt {{{16}^2} - {{14}^2}} = 2\sqrt {15} \\
\Rightarrow AH = \dfrac{{AB.AC}}{{BC}} = \dfrac{{7\sqrt {15} }}{4}\\
\Rightarrow BH = x = \sqrt {A{B^2} - A{H^2}} = \dfrac{{15}}{4}\\
\Rightarrow y = 16 - x = \dfrac{{49}}{4}\\
\text{Vậy}\,x = \dfrac{{15}}{4};y = \dfrac{{49}}{4}\\
B2)\\
\dfrac{1}{{A{H^2}}} = \dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}}\\
\Rightarrow AC = 8\\
\Rightarrow {S_{ABC}} = \dfrac{1}{2}.AB.AC = \dfrac{1}{2}.6.8 = 24\left( {c{m^2}} \right)
\end{array}$
