Rezolvatii ecuatiile:

Licuricy Mar Aug 09, 2011 5:03 pm

a).x!/(x-5)!-3*x!/(x-4)!=21*x!/(x-3)!
b).6*x!/(x-1)!*1!+6*(x+2)!/(x+2-3)!*3!=13*(x+1)!/(x+1-2)!*2!
c).(x-2)!/2!+x!/(x-2)!*2!=41
Este vorba despre ecuatii cu aranjamente si combinari

**Admin** Mar Aug 09, 2011 8:15 pm

a).x!/(x-5)!-3*x!/(x-4)!=21*x!/(x-3)!

Rezolvare:
$\begin{array}{l} \frac{{x!}}{{\left( {x - 5} \right)!}} - 3 \cdot \frac{{x!}}{{\left( {x - 4} \right)!}} = 21 \cdot \frac{{x!}}{{\left( {x - 3} \right)!}}\left| { \cdot \left[ {\frac{{\left( {x - 3} \right)!}}{{x!}}} \right]} \right. \Rightarrow \\ \left( {x - 3} \right)\left( {x - 4} \right) - 3 \cdot \left( {x - 3} \right) = 21 \Rightarrow x = 10 \end{array}$

**Admin** Mar Aug 09, 2011 8:35 pm

b).6*x!/(x-1)!*1!+6*(x+2)!/(x+2-3)!*3!=13*(x+1)!/(x+1-2)!*2!

Voi rezolva problema luând în considerare formulele de la combinări :

$\begin{array}{l} 6 \cdot \frac{{x!}}{{\left( {x - 1} \right)! \cdot 1!}} + \frac{{6 \cdot \left( {x + 2} \right)!}}{{\left( {x + 2 - 3} \right)! \cdot 3!}} = 13 \cdot \frac{{\left( {x + 1} \right)!}}{{\left( {x + 1 - 2} \right)! \cdot 2!}}\\ 6 \cdot C_x^1 + 6 \cdot C_{x + 2}^3 = 13C_{x + 1}^2 \Leftrightarrow 6 \cdot C_x^1 - 13C_{x + 1}^2 + 6 \cdot C_{x + 2}^3 = 0 \end{array}$

Acum vom folosi aceasta formula consecutiv:

$C_n^p = C_{n - 1}^p - C_{n - 1}^{p - 1}$
$\left\{ \begin{array}{l} C_{x + 1}^2 = C_x^2 - C_x^1\\ C_{x + 2}^3 = C_{x + 1}^3 - C_{x + 1}^2 = C_x^3 - C_x^2 - \left( {C_x^2 - C_x^1} \right) \end{array} \right.$

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