Calcule :(

Natalia Lun Feb 20, 2012 12:35 am

Buna seara!
$( \frac{x}{\sqrt{3}} -\frac{1}{\sqrt{2}}) ( \frac{x}{\sqrt{3}}+ \frac{1}{\sqrt{2}})=$
$( x\sqrt{3}-x\sqrt{2}+1) ( x\sqrt{3}-x\sqrt{2}+1)=$

**Admin** Lun Feb 20, 2012 3:19 am

Natalia a scris:Problema 1

Vom folosi bine cunoscuta formula : $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ , unde :
$\left\{ \begin{array}{l} a = \frac{x}{{\sqrt 3 }}\\ b = \frac{1}{{\sqrt 2 }} \end{array} \right. \Rightarrow \left( {\frac{x}{{\sqrt 3 }} - \frac{1}{{\sqrt 2 }}} \right)\left( {\frac{x}{{\sqrt 3 }} + \frac{1}{{\sqrt 2 }}} \right) = {\left( {\frac{x}{{\sqrt 3 }}} \right)^2} - {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = \frac{{{x^2}}}{3} - \frac{1}{2} = \frac{{2{x^2} - 3}}{6}$

**Admin** Lun Feb 20, 2012 3:20 am

Natalia a scris:Problema 2

Analog rezolvării problemei 1.

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