Sa se rezolve ecuatiile:

crystynaz Dum Oct 02, 2011 7:12 pm

1)Sa se rezolve ecuatiile:

$a) \left [ \frac{x-2}{3} \right ]=\frac{x+2}{4}$

$b) \left [ \frac{x+1}{2} \right ]=\frac{3x-1}{4}$

2) Sa se rezolve ecuatia:

$\left [ 2x+1 \right ]=\left [ 3x-1 \right ]$

**Admin** Mar Oct 04, 2011 6:41 am

crystynaz a scris: $a) \left [ \frac{x-2}{3} \right ]=\frac{x+2}{4}$

Voi arata modelul de calcul pentru aceasta problema ținând cont de faptul ca sunt cunoscute formulele elementare din capitolul "parte întreagă", prin alte cuvinte nu voi explica pas cu pas soluția prezentată.

$\begin{array}{l} \left[ {\frac{{x - 2}}{3}} \right] = \frac{{x + 2}}{4}\\ {\text{Notam }}k = \left[ {\frac{{x - 2}}{3}} \right],k \in Z \Rightarrow \left\{ \begin{array}{l} \frac{{x + 2}}{4} = k \Rightarrow 4k = x + 2 \Rightarrow x = 4k - 2{\rm{ }}\left( {{\text{relatia 1}}} \right)\\ \left[ {\frac{{x - 2}}{3}} \right] = k\,\,\,\,\left( {{\text{relatia 2}}} \right) \end{array} \right.\\ \left[ {\frac{{x - 2}}{3}} \right] = \frac{{x + 2}}{4}\mathop \Leftrightarrow \limits^{{\rm{relatia 1}} + {\rm{relatia 2}}} \left[ {\frac{{4k - 4}}{3}} \right] = k \Leftrightarrow k \le \frac{{4k - 4}}{3} < k + 1\\ \left\{ \begin{array}{l} \frac{{4k - 4}}{3} < k + 1\\ k \le \frac{{4k - 4}}{3} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 4k - 4 < 3k + 3 \Rightarrow k < 7\\ 3k \le 4k - 4 \Rightarrow 4 \le k \end{array} \right.,\,\,k \in Z \Rightarrow k \in \left\{ {4,5,6} \right\}\\ x = 4k - 2 \Rightarrow \left\{ \begin{array}{l} x = 14\\ x = 18\\ x = 22 \end{array} \right. \end{array}$

**Admin** Mar Oct 04, 2011 6:43 am

crystynaz a scris: $b) \left [ \frac{x+1}{2} \right ]=\frac{3x-1}{4}$

In mod analog problemei a.

**Greatmath** Mar Oct 04, 2011 7:28 am

$\left [ 2x+1 \right ]=\left [ 3x-1 \right ]$

Cea mai eficienta soluție ar fi:
$\begin{array}{l} \left[ {2x + 1} \right] = \left[ {3x - 1} \right]\\ {\text{Notam }}\left[ {2x + 1} \right] = \left[ {3x - 1} \right] = k,\,\,k \in Z \Rightarrow \left\{ \begin{array}{l} \left[ {3x - 1} \right] = k\\ \left[ {2x + 1} \right] = k \end{array} \right.\\ \left[ {3x - 1} \right] = k \Leftrightarrow k \le 3x - 1 < k + 1 \Leftrightarrow \frac{{k + 1}}{3} \le x < \frac{{k + 2}}{3}\\ \left[ {2x + 1} \right] = k \Leftrightarrow k \le 2x + 1 < k + 1 \Leftrightarrow \frac{{k - 1}}{2} \le x < \frac{k}{2}\\ {\text{Combinand cele 2 ecuatii vom obtine:}}\\ \max \left\{ {\frac{{k + 1}}{3},\frac{{k - 1}}{2}} \right\} \le x < \min \left\{ {\frac{{k + 2}}{3},\frac{k}{2}} \right\}\\ {\text{De exemplu, pentru }}k = 4 \Rightarrow \max \left\{ {\frac{5}{3}, - \frac{3}{2}} \right\} \le x < \min \left\{ {2,2} \right\} \Rightarrow x \in \left[ {\frac{5}{3},2} \right)\\ {\text{In concluzie, ecuatia are o infinitate de solutii}}{\text{.}} \end{array}$

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