Question

Se $x = 3 - \sqrt{3} + \frac{1}{3+\sqrt{3}} - \frac{1}{\sqrt{3}-3}$ então:

Resposta: $1 \leq x < 3$

Answer 1

$x=3-\sqrt3+\frac{1}{3+\sqrt{3}}-\frac{1}{\sqrt{3}-3}=3-\sqrt3+\frac{1}{3+\sqrt{3}}\cdot\frac{3-\sqrt{3}}{3-\sqrt{3}}-\frac{1}{\sqrt{3}-3}\cdot\frac{\sqrt{3}+3}{\sqrt{3}+3}=\\\\ =3-\sqrt3+\frac{3-\sqrt{3}}{9-3}-\frac{\sqrt{3}+3}{3-9}=3-\sqrt3+\frac{3-\sqrt{3}}{6}-\frac{\sqrt{3}+3}{-6}=$

$=3-\sqrt3+\frac{3-\sqrt{3}}{6}+\frac{\sqrt{3}+3}{6}=3-\sqrt3+\frac{3-\sqrt{\not3}+\sqrt{\not3}+3}{6}=\\\\ =3-\sqrt3+\frac66=3-\sqrt3+1=4-\sqrt3$

Como $1<\sqrt3<2,$ temos que:

$-1>-\sqrt3>-2\Rightarrow\\\\-2<-\sqrt3<-1\Rightarrow\\\\4-2<4-\sqrt3<4-1\Rightarrow\\\\ 2<4-\sqrt3<3\Rightarrow\\\\ 2<x<3\Rightarrow\\\\ \boxed{1 \leq x<3}\text{ (c.q.d.)}$