Đáp án:
$\begin{array}{l}
3)a)\\
Theo\,Pytago:\\
{y^2} = {7^2} + {9^2} = 130\\
\Leftrightarrow y = \sqrt {130} \\
S = \dfrac{1}{2}.7.9 = \dfrac{1}{2}.x.y\\
\Leftrightarrow x = \dfrac{{63}}{y} = \dfrac{{63}}{{\sqrt {130} }} = \dfrac{{63\sqrt {130} }}{{130}}\\
b)Theo\,Pytago:\\
\left\{ \begin{array}{l}
{y^2} + {y^2} = {\left( {x + x} \right)^2}\\
{5^2} + {x^2} = {y^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2{y^2} = 4{x^2}\\
{y^2} = {x^2} + 25
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{y^2} = 2{x^2}\\
2{x^2} = {x^2} + 25
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = \sqrt 2 x\\
{x^2} = 25 \Leftrightarrow x = 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 5\sqrt 2 \\
x = 5
\end{array} \right.\\
Vậy\,x = 5;y = 5\sqrt 2 \\
4)a)\\
{3^2} = 2.x\\
\Leftrightarrow x = \dfrac{9}{2}\\
Theo\,Pytago;\\
{3^2} + {x^2} = {y^2}\\
\Leftrightarrow {y^2} = 9 + {\left( {\dfrac{9}{2}} \right)^2} = \dfrac{{117}}{4}\\
\Leftrightarrow y = \dfrac{{3\sqrt {13} }}{2}\\
Vậy\,x = \dfrac{9}{2};y = \dfrac{{3\sqrt {13} }}{2}\\
b)\dfrac{{AB}}{{AC}} = \dfrac{3}{4}\\
\Leftrightarrow AC = \dfrac{{4AB}}{3} = \dfrac{{4.15}}{3} = 20\\
B{C^2} = A{B^2} + A{C^2}\\
= {15^2} + {20^2} = 625\\
\Leftrightarrow BC = 25 = y\\
{S_{ABC}} = \dfrac{1}{2}.15.20 = \dfrac{1}{2}.x.y\\
\Leftrightarrow x = \dfrac{{15.20}}{{25}} = 12\\
Vậy\,x = 12;y = 25
\end{array}$
