Question

Repede va rog....o sa dau test....

Answer 1

   
[tex]\displaystyle\\ \begin{cases} x \sqrt{3-2 \sqrt{2}}+y\sqrt{7-4\sqrt{3}}=x \sqrt{2} -y \sqrt{3} +4\\\\ x \sqrt{4-2 \sqrt{3}}+y\sqrt{3+2\sqrt{2}}=x \sqrt{3} +y \sqrt{2} +1 \end{cases}\\\\\\ \begin{cases} x \sqrt{2+1-2 \sqrt{2}}+y\sqrt{4+3-4\sqrt{3}}=x \sqrt{2} -y \sqrt{3} +4\\\\ x \sqrt{3+1-2 \sqrt{3}}+y\sqrt{2+1+2\sqrt{2}}=x \sqrt{3} +y \sqrt{2} +1 \end{cases}\\\\\\ [/tex]


[tex]\displaystyle\\ \begin{cases} x \sqrt{2-2 \sqrt{2}+1}+y\sqrt{4-4\sqrt{3}+3}=x \sqrt{2}-y\sqrt{3}+4\\\\ x \sqrt{3-2\sqrt{3}+1}+y\sqrt{2+2\sqrt{2}+1}=x\sqrt{3} +y \sqrt{2}+1 \end{cases}\\\\\\ \begin{cases} x\sqrt{( \sqrt{2}-1)^2}+y\sqrt{(2-\sqrt{3})^2}=x \sqrt{2}-y\sqrt{3}+4\\\\ x\sqrt{(\sqrt{3}-1)^2}+y\sqrt{(\sqrt{2}+1)^2}=x\sqrt{3}+y\sqrt{2}+1 \end{cases}[/tex]


[tex]\displaystyle\\ \begin{cases} x\sqrt{( \sqrt{2}-1)^2}+y\sqrt{(2-\sqrt{3})^2}=x \sqrt{2}-y\sqrt{3}+4\\\\ x\sqrt{(\sqrt{3}-1)^2}+y\sqrt{(\sqrt{2}+1)^2}=x\sqrt{3}+y\sqrt{2}+1 \end{cases}\\\\\\ \begin{cases} x(\sqrt{2}-1)+y(2-\sqrt{3})=x \sqrt{2}-y\sqrt{3}+4\\\\ x(\sqrt{3}-1)+y(\sqrt{2}+1)=x\sqrt{3}+y\sqrt{2}+1 \end{cases}\\\\\\ \begin{cases} x\sqrt{2}-x+2y-y\sqrt{3}=x \sqrt{2}-y\sqrt{3}+4\\\\ x\sqrt{3}-x+y\sqrt{2}+y=x\sqrt{3}+y\sqrt{2}+1 \end{cases}[/tex]


[tex]\displaystyle\\ \begin{cases} x\sqrt{2}-x\sqrt{2} -x+2y-y\sqrt{3}+y\sqrt{3}=4\\\\ x\sqrt{3}-x\sqrt{3}-x+y\sqrt{2}-y\sqrt{2}+y=1 \end{cases}\\\\\\ \begin{cases} -x+2y=4\\\\ -x+y=1 \end{cases} ~~~~~\text{Din ecuatia 1 scadem ecuatia 2.}\\\\\\ ~~~~~/~~~~~\boxed{y =3}\\\\ -x+y = 1\\\\ -x+3 = 1\\\\ x = 3-1\\\\ \boxed{x = 2}[/tex]