Tìm số nguyên x, biết:
\(4\frac{1}{3}\left( {\frac{1}{6} - \frac{1}{2}} \right) \le x \le \frac{2}{3}\left( {\frac{1}{3} - \frac{1}{2} - \frac{3}{4}} \right)\)
Ta có:
\(\begin{array}{l}
4\frac{1}{3}\left( {\frac{1}{6} - \frac{1}{2}} \right) \le x \le \frac{2}{3}\left( {\frac{1}{3} - \frac{1}{2} - \frac{3}{4}} \right)\\
\Leftrightarrow \frac{{13}}{3}.\frac{{ - 1}}{3} \le x \le \frac{2}{3}.\frac{{ - 11}}{{12}}\\
\Leftrightarrow \frac{{ - 13}}{9} \le x \le \frac{{ - 11}}{{18}}
\end{array}\)
Mà x là số nguyên nên suy ra x = -1
-- Mod Toán 6
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