Question

3. Se consideră numerele reale a = (2p) a) Arată că a 11 2 √√24 √32-4-(√2-√18) √3 şi b= 1 1 + 1-2 (3p) b) Arată că numărul N=2(a+b) aparţine intervalului (2,√7). 1 1 + + 2-3 3-4 4-5​

Answer 1

a)

$a = \dfrac{ \sqrt{24} }{ \sqrt{32} - 4 \cdot (\sqrt{2} - \sqrt{18})} \cdot \sqrt{3} =$

$= \dfrac{2\sqrt{6} \cdot \sqrt{3} }{4\sqrt{2} - 4 \cdot (\sqrt{2} - 3\sqrt{2})}$

$= \dfrac{2\sqrt{18} }{4\sqrt{2} - 4\sqrt{2} + 12\sqrt{2}}$

$= \dfrac{6\sqrt{2} }{12\sqrt{2}} = \bf \dfrac{1}{2}$

b)

$b = \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \dfrac{1}{4 \cdot 5} =$

$= \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5}$

$= \dfrac{1}{1} - \dfrac{1}{5} = \dfrac{5 - 1}{5} = \bf \dfrac{4}{5}$

$N = 2(a+b) = 2\bigg(\dfrac{^{5)} 1}{2} + \dfrac{^{2)} 4}{5}\bigg) = 2 \cdot \dfrac{5 + 8}{10} = \bf \dfrac{13}{5}$

$\dfrac{13}{5} = 2\dfrac{3}{5} > 2 \iff N > 2$

$\sqrt{7} = \dfrac{5 \sqrt{7} }{5} = \dfrac{\sqrt{175} }{5} > \dfrac{\sqrt{169} }{5} = \dfrac{13}{5}$

$\dfrac{13}{5} < \sqrt{7} \iff N < \sqrt{7}$

$2 < N < \sqrt{7} \implies \bf N \in \big(2; \sqrt{7} \big)$

q.e.d.