Oznaczmy:

x - pierwsza z liczb

y - druga z liczb

 

Z treści zadania wiemy, że:

$\{\begin{array}{l}x+y=\sqrt{2}\\ x-y=\sqrt{3}\end{array}$  

$\{\begin{array}{l}x=\sqrt{2}-y\\ \sqrt{2}-y-y=\sqrt{3}\end{array}$  

$\{\begin{array}{l}x=\sqrt{2}-y\\ -2y=\sqrt{3}-\sqrt{2}\end{array}$  

$\{\begin{array}{l}x=\sqrt{2}-y\\ y=\frac{\sqrt{3}-\sqrt{2}}{-2}\end{array}$  

$\{\begin{array}{l}x=\sqrt{2}-y\\ y=\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\end{array}$  

$\{\begin{array}{l}x=\sqrt{2}-\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\\ y=\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\end{array}$  

$\{\begin{array}{l}x=\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\\ y=\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\end{array}$  

Obliczmy iloczyn tych liczb.

$x\cdot y=\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\right)\cdot \left(\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\right)=$  

$={\left(\frac{\sqrt{2}}{2}\right)}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}$  

 

Odp. D