Răspuns:
2. [tex]a=\sqrt{81}-3\sqrt{3}+\sqrt{27}=9-3\sqrt{3}+3\sqrt{3}=9[/tex]
[tex]b=|2-\sqrt{3}| + \frac{1}{2-\sqrt{3}} = 2-\sqrt{3}+\frac{2+\sqrt{3}}{2^2-\sqrt{3}^2}=2-\sqrt 3 +\frac{2+\sqrt 3}{4-3}=2-\sqrt 3 + 2+\sqrt 3 =4.[/tex]
Atunci media geometrica este: [tex]m=\sqrt{a\cdot b}=\sqrt{36}=6[/tex].
3. a)[tex]E(-1)=8\cdot(-1)^3 + 4\cdot(-1)^2 = 8\cdot (-1) + 4\cdo 1 =-8+4=-4. [/tex]
[tex]F(\frac{1}{3})=9\cdot (\frac{1}{3})^2-12\cdot \frac{1}{3}+4 = 9\cdot \frac{1}{9} - 4+4= 1-4+4=1.[/tex]
Rezulta ca: [tex] E(-1)-F(\frac{1}{3})= -4-1=-5.[/tex]
b) [tex]E(x)=8x^3+4x^2=4x^2\cdot (2x+1)[/tex]
[tex]F(x)=9x^2-12x+4=(3x)^2 - 2\cdot (3x)\cdot 2 + 2^2 = (3x-2)^2[/tex]
Am folosit formula: [tex] (a-b)^2=a^2-2ab+b^2 [/tex].
