Ecuatie de gr. I

Stefy Sam Sept 15, 2012 2:08 am

1.Sa se rezolve ecuatiile: $(3-\sqrt{5}x)(-4)-(2x-7)\sqrt{5}-18= - \sqrt{3}\cdot (\sqrt{27}-\sqrt{12}x)$
2.Sa se determine $x \in \mathbb{Z}$ pentru care numarul:
$A=\frac{\sqrt{(3-\sqrt{5})^{2}}+\sqrt{(4+\sqrt{5})^{2}}-6+x}{x} \in \mathbb{Z}$
Ma poate ajuta cineva?
Multumesc! Smile

**Admin** Sam Sept 15, 2012 5:00 am

Problema 1

Calculul amănunțit ar fi :

$\begin{array}{l} - 4 \times \left( {3 - \sqrt 5 x} \right) - \sqrt 5 \left( {2x - 7} \right) - 18 = - \sqrt 3 \left( {\sqrt {27} - \sqrt {12} x} \right)\\ - 12 + 4\sqrt 5 x - 2\sqrt 5 x + 7\sqrt 5 - 18 = - \sqrt {3 \times 27} + 6x\\ - 30 + 2\sqrt 5 x + 7\sqrt 5 = - 9 + 6x \Leftrightarrow 2\sqrt 5 x - 6x = 21 - 7\sqrt 5 \\ x\left( {2\sqrt 5 - 6} \right) = 21 - 7\sqrt 5 \Rightarrow x = \frac{{21 - 7\sqrt 5 }}{{2\sqrt 5 - 6}} = \frac{{7\left( {3 - \sqrt 5 } \right)}}{{ - 2\left( {3 - \sqrt 5 } \right)}} = - \frac{7}{2} \end{array}$

**Admin** Sam Sept 15, 2012 5:05 am

Problema 2

$\begin{array}{l} A = \frac{{\sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} + \sqrt {{{\left( {4 + \sqrt 5 } \right)}^2}} - 6 + x}}{x} = \frac{{3 - \sqrt 5 + 4 + \sqrt 5 - 6 + x}}{x} = \frac{{x + 1}}{x} = 1 + \frac{1}{x} \in Z\\ \Leftrightarrow x = \pm 1 \end{array}$

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