ecuatii

ali · Mesaj de **ali** » 12 Noi 2008, 01:25

oana16 scrie: $\left[ {\frac{{x + 13}}{6}} \right] = \frac{{x + 20}}{7} \]$

$\begin{array}{l} \left[ {\frac{{x + 13}}{6}} \right] = \frac{{x + 20}}{7} \\ Notam:\left[ {\frac{{x + 13}}{6}} \right] = k,k \in Z,at: \\ k = \frac{{x + 20}}{7} \Rightarrow x = 7k - 20 \\ Acum:k \to x \Rightarrow \\ \left[ {\frac{{7k - 20 + 13}}{6}} \right] = k \Leftrightarrow \left[ {\frac{{7k - 7}}{6}} \right] = k \\ {\rm Aceasta egalitate este egala cu dubla inegalitate:} \\ k \le \frac{{7k - 7}}{6} < k + 1 \\ {\mathop{\rm Re}\nolimits} zolvand{\rm inegalitatiile se obtine ca solutie:} \\ {\rm x} \in \left\{ {{\rm 36,43,50,57,64}} \right\} \\ \end{array}$
Primul exercitiu se rezolva asemanator

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