puteri

inc.nci Joi Oct 11, 2012 3:12 am

$x=\sqrt[4^{n}-1]{\left ( \sqrt{a}\times \sqrt[4]{a} \times ...\times \sqrt[2^{n}]{a}\right )^{2^{n}}}$

$y=\sqrt[n]{\left ( \sqrt{a}\times \sqrt[6]{a}\times ...\times \sqrt[n^{2}+n]{a} \right )^{n+1}}$

Precizati daca:
a)x>y
b) $x^{2^{n}+1}=y$
c)xy<1

PS:Am ajuns la faptul ca $x=\sqrt{a}^{\tfrac{2^{n+1}}{2^{n}+1}}$ $y=\sqrt{a}^{\left ( n+1 \right )^{3}}$
Iar cand a venit vorba de comparare m-am impotmolit...

**Admin** Joi Oct 11, 2012 7:43 pm

Voi folosi proprietara: ${x^n} \times {x^m} = {x^{n + m}}$ . Pentru x:
$\begin{array}{l} x = \sqrt[{{4^n} - 1}]{{{{\left( {\underbrace {\sqrt a \times \sqrt[4]{a} \times ... \times \sqrt[{{2^n}}]{a}}_{ = P}} \right)}^{{2^n}}}}}\\ \left\{ \begin{array}{l} P = \sqrt a \times \sqrt[4]{a} \times ... \times \sqrt[{{2^n}}]{a} = {a^{\frac{1}{2}}} \times {a^{\frac{1}{4}}} \times ... \times {a^{\frac{1}{{{2^n}}}}} = {a^{\frac{1}{2} + ... + \frac{1}{{{2^n}}}}} = {a^{1 - \frac{1}{{{2^n}}}}} = {a^{\frac{{{2^n} - 1}}{{{2^n}}}}}\\ \frac{1}{2} + ... + \frac{1}{{{2^n}}} = \frac{1}{2}\left( {\underbrace {1 + \frac{1}{2} + ... + \frac{1}{{{2^{n - 1}}}}}_{{\text{Progresie geometrica}}}} \right) = \frac{1}{2} \times \frac{{1 - \frac{1}{{{2^n}}}}}{{1 - \frac{1}{2}}} = 1 - \frac{1}{{{2^n}}} \end{array} \right. \end{array}$
$\left\{ \begin{array}{l} x = \sqrt[{{4^n} - 1}]{{{{\left( {{a^{\frac{{{2^n} - 1}}{{{2^n}}}}}} \right)}^{2n}}}} = \sqrt[{{4^n} - 1}]{{{a^{{2^n} \times \frac{{{2^n} - 1}}{{{2^n}}}}}}} = {a^{\frac{{{2^n} - 1}}{{{4^n} - 1}}}} = {a^{\frac{1}{{{2^n} + 1}}}}\\ \frac{{{2^n} - 1}}{{{4^n} - 1}} = \frac{{{2^n} - 1}}{{{2^{2n}} - 1}} = \frac{{{2^n} - 1}}{{\left( {{2^n} - 1} \right)\left( {{2^n} + 1} \right)}} = \frac{1}{{{2^n} + 1}} \end{array} \right.$

Pentru y se procedează la fel .........

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