Ex2
1) Ф≈1,618
2) [tex](\frac{1+\sqrt{5}}{2})^{2}=\frac{1+2\sqrt{5}+5}{4}=\frac{6+2\sqrt{5}}{4} [/tex]
[tex](\frac{1+\sqrt{5}}{2})^{2}=\frac{3+\sqrt{5}}{2}=\frac{1+2+\sqrt{5}}{2}=\frac{1+\sqrt{5}}{2}+1 [/tex]
Donc Ф²=Ф+1
Ex3
a)
[tex](3-\sqrt{2})(3+\sqrt{2})=3^{2}-2=9-2=7 [/tex]
b)
[tex](3-\sqrt{2})(3+\sqrt{2})=7 [/tex]
⇔[tex](3+\sqrt{2})=\frac{7}{(3-\sqrt{2})} [/tex]
⇔[tex] \frac{(3+\sqrt{2})}{7}=\frac{1}{(3-\sqrt{2})} [/tex]
⇔[tex]\frac{4(3+\sqrt{2})}{7}=\frac{4}{(3-\sqrt{2})} [/tex]
