Question

Sa se calculeze produsul

Answer 1

$x^{\frac{1}{\sqrt{1}+\sqrt{2}}}*x^{\frac{1}{\sqrt{2}+\sqrt{3}}}*x^{\frac{1}{ \sqrt{3}+\sqrt{4}}}*...*x^{\frac{1}{\sqrt{99}+\sqrt{100}}}*x^{\frac{1}{\sqrt{100}+\sqrt{101}}}=$

amplificam puterile:prima cu $\sqrt{1}-\sqrt{2}$,
 a doua cu $\sqrt{2}-\sqrt{3}$
 si tot asa,
iar ulima cu $\sqrt{100}-\sqrt{101}$

Si obtinem:

$=x^{\frac{\sqrt{1}-\sqrt{2}}{1-2}}}*x^{\frac{\sqrt{2}-\sqrt{3}}{2-3}}*...*x^{\frac{\sqrt{100}-\sqrt{101}}{100-101}}=$

$\sqrt{2}>\sqrt{1}$
$\sqrt{3}>\sqrt{2}$
............................................................
$\sqrt{101}>\sqrt{100}$

$=x^{\sqrt{2}-\sqrt{1}}*x^{\sqrt{3}-\sqrt{12}}*...*x^{\sqrt{101}-\sqrt{100}}*=$

$=x^{(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+....+(\sqrt{101}-\sqrt{100})}=$

$=x^{(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+....+\sqrt{101}-\sqrt{100}}=$

$=x^{-\sqrt{1}+\sqrt{101}}=$




($x=(\sqrt[4]{\sqrt[5]{3}})^{\sqrt{101}+1}$

$x=(\sqrt[20]{3})^{\sqrt{101}+1}$

$x=({3}^{\frac{1}{20}})^{\sqrt{101}+1}$

$x=3^{\frac{\sqrt{101}+1}{20}}$        )




$=(3^{\sqrt{101}+1})^{\frac{-\sqrt{1}+\sqrt{101}}{20}}=$

$=3^{(\sqrt{101}+1)(\frac{-\sqrt{1}+\sqrt{101}}{20})}=$

$=3^{\frac{(\sqrt{101}+1)(\sqrt{101}-\sqrt{1})}{20}}=$

$=3^{\frac{101-1}{20}}=$

$=3^{\frac{100}{20}}=$

$=3^{5}=$

$\boxed{=243}$


Cateva formule o sa-ti prinda bine:

$\boxed{\sqrt[n]{a}=a^{\frac{1}{n}}}$

$\boxed{\sqrt[m]{\sqrt[n]{a}}=\sqrt[m*n]{a}=a^{\frac{1}{m*n}}}$