Cum se rezolve ecuatia :
\( [\frac{x-2}{3}]=[\frac{x+1}{2}] \) ?
Parte intreaga
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mihai miculita
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\( \[\frac{x-2}{3}\]=\[\frac{x+1}{2}\]=k\in\mathbb{Z}\Leftrightarrow\left\{\begin{array}{c}k\le\frac{x-2}{3}<k+1\\
k\le\frac{x+1}{2}<k+1\end{array} \Leftrightarrow \left\{\begin{array}{c}3k+2\le x<3k+5\\
2k-1\le x<2k+1\end{array} \Leftrightarrow x\in[2k-1;2k+1)\cap[3k+2;3k+5);k\in\mathbb{Z}\Rightarrow x\in [-11;-10)\cup[-9;-5)\cup [-4;-3). \)
REMARCA:
\( \mbox{Partea intreaga a numarului } x\in \mathbb{R} \mbox{, este cel mai mare numar intreg care nu depaseste acel numar real.} \)
\( \mbox{Cu alte cuvinte, avem: } [x]=k\in \mathbb{Z}\Leftrightarrow k\le x <k+1. \)
k\le\frac{x+1}{2}<k+1\end{array} \Leftrightarrow \left\{\begin{array}{c}3k+2\le x<3k+5\\
2k-1\le x<2k+1\end{array} \Leftrightarrow x\in[2k-1;2k+1)\cap[3k+2;3k+5);k\in\mathbb{Z}\Rightarrow x\in [-11;-10)\cup[-9;-5)\cup [-4;-3). \)
REMARCA:
\( \mbox{Partea intreaga a numarului } x\in \mathbb{R} \mbox{, este cel mai mare numar intreg care nu depaseste acel numar real.} \)
\( \mbox{Cu alte cuvinte, avem: } [x]=k\in \mathbb{Z}\Leftrightarrow k\le x <k+1. \)