Question

va rog foarte mult ms anticipat

Answer 1

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]