Question

Fie numarul a=1/(1+radical 3)+1/(3-2 radical 3)+1/(2-radical 3).
a) Scrieti numarul a sub o forma cat mai simpla.
b)Determinați numerele naturale mai mici decat a.
c)Calculati (2a+11/a) la puterea -1

Answer 1

$a) \ \ a=\frac{1}{1+\sqrt{3}}^{(1-\sqrt{3}}+\frac{1}{3-2\sqrt{3}}^{(3+2\sqrt{3}}+ \frac{1}{2-\sqrt{3}}^{(2+\sqrt{3}} \\\\ a= \frac{1-\sqrt{3}}{1-3}+\frac{3+2\sqrt{3}}{9-12}+ \frac{2+\sqrt{3}}{4-3} \\\\ a= -\frac{1-\sqrt{3}}{2}^{(3}-\frac{3+2\sqrt{3}}{3}}^{(2}+ \frac{2+\sqrt{3}}{1}^{(6} \\\\ a=-3+3\sqrt{3}-6-4\sqrt{3}+12+6\sqrt{3} \\\\ \boxed{a= 5\sqrt{3} +3} \\\\\\\\ b)\underline{\sqrt{3} \approx 1,73} \\\\ a= 5*1,73+3 \\\\ a= 8,65+3 \\\\ \underline{\underline{a \cong11,65}}$
$\hbox{Numerele naturale mai mici decat a sunt: } \\\\ N \ \textless \  11,65 \\\\ \boxed{\boxed{N \in \{1;2;3;4;5;6;7;8;9;10;11 \}}} \\\\\\\\ c) (\frac{2a+11}{a})^{-1} = \frac{1}{\frac{2a+11}{a}} \longrightarrow \frac{a}{2a+11} \\\\\\ \frac{a}{2a+11}= \frac{5\sqrt{3}+3}{2(5\sqrt{3}+3)+11}= \frac{5\sqrt{3}+3}{10\sqrt{3}+6+11}= \frac{5\sqrt{3}+3}{10\sqrt{3}+17}^{(10\sqrt{3}-17} \\\\\\ \frac{a}{2a+1}=\frac{(5\sqrt{3}+3)(10\sqrt{3}-17)}{300-289}= \frac{ 150-85\sqrt{3}+30\sqrt{3}-51}{11}$
$\frac{a}{2a+11} = \frac{99-55\sqrt{3}}{11} = \frac{\not 11(9-5\sqrt{3})}{\not11} \\\\\\ \boxed{\frac{a}{2a+11} \longrightarrow 9-5\sqrt{3}}$