EXERCITIUL 5
[tex]\frac{\sqrt{5}}{2-\sqrt{5}}+\frac{2}{2+\sqrt{5}}
\\\text{(Aducem la acelasi numitor)}
\\\frac{\sqrt{5}(2+\sqrt{5}) + 2(2-\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})} =
\\\frac{2\sqrt{5}+5 + 4-2\sqrt{5}}{4-5} = -9
[/tex]
EXERCITIUL 6
[tex]2x^2 + 5x-3=0
\\\Delta = 25+24=49
\\x_1 = \frac{-5+7}{4} = \frac{1}{2}
\\x_2 = \frac{-5-7}{4} = -3
\\A = \{-3, \frac{1}{2}\}
\\A-\{-3,2\} = \{\frac{1}{2}\}[/tex]
EXERCITIUL 7
[tex]AD=BC=5\\AB=DC=3
\\Conform\ TP: AD^2=AB^2+BD^2
\\25=9+BD^2\Rightarrow BD=4
\\A_{ABCD}=A_{ADB}+A_{BDC}
\\A_{ADB} = \frac{AB\cdot BD}{2}=\frac{12}{2}=6
\\A_{BDC}=\sqrt{p(p-BD)(p-BC)(p-DC)}
\\p=\frac{BD+BC+DC}{2}=\frac{12}{2}=6
\\A_{BDC}=\sqrt{6(6-4)(6-3)(6-5)}=\sqrt{6\cdot3\cdot2}=\sqrt{36}=6
\\A_{ABCD}=6+6=12[/tex]
