Tìm các số nguyên \(x\) biết:a) \(\frac{1}{3}+\fra...

B

- Hướng dẫn giải

Phương pháp giải:

Tính giá trị của từng về rồi so sánh để tìm các số nguyên \(x\).

Giải chi tiết:

a) Ta có:

\(\begin{array}{l} + )\frac{1}{3} + \frac{{ - 2}}{5} + \frac{1}{6} + \frac{{ - 1}}{5}\\ = \left( {\frac{1}{3} + \frac{1}{6}} \right) + \left( {\frac{{ - 2}}{5} + \frac{{ - 1}}{5}} \right)\\ = \left( {\frac{2}{6} + \frac{1}{6}} \right) + \frac{{ - 3}}{5}\\ = \frac{3}{6} + \frac{{ - 3}}{5} = \frac{1}{2} + \frac{{ - 3}}{5}\\ = \frac{5}{{10}} + \frac{{ - 6}}{{10}} = \frac{{ - 1}}{{10}};\end{array}\)

 \(\begin{array}{l} + )\frac{{ - 3}}{4} + \frac{2}{7} + \frac{{ - 1}}{4} + \frac{3}{5} + \frac{5}{7}\\ = \left( {\frac{{ - 3}}{4} + \frac{{ - 1}}{4}} \right) + \left( {\frac{2}{7} + \frac{5}{7}} \right) + \frac{3}{5}\\ = \frac{{ - 4}}{4} + \frac{7}{7} + \frac{3}{5}\\ = ( - 1) + 1 + \frac{3}{5}\\ = 0 + \frac{3}{5} = \frac{3}{5}.\end{array}\)

Do đó  \(\frac{-1}{10}\le \frac{x}{10}<\frac{3}{5}\) hay \(\frac{-1}{10}\le \frac{x}{10}<\frac{6}{10}\), từ đó suy ra \(-1\le x<6\)

Vậy \(x\in \text{ }\!\!\{\!\!\text{ }-1;\,\,0;\,\,1;\,\,2;\,\,3;\,\,4;\,\,5\}\)

b) Ta có:

\(\begin{array}{l} + )\frac{5}{{17}} + \frac{{ - 4}}{9} + \frac{{ - 20}}{{31}} + \frac{{12}}{{17}} + \frac{{ - 11}}{{31}}\\ = \left( {\frac{5}{{17}} + \frac{{12}}{{17}}} \right) + \left( {\frac{{ - 20}}{{31}} + \frac{{ - 11}}{{31}}} \right) + \frac{{ - 4}}{9}\\ = \frac{{17}}{{17}} + \frac{{ - 31}}{{31}} + \frac{{ - 4}}{9}\\ = 1 + ( - 1) + \frac{{ - 4}}{9}\\ = 0 + \frac{{ - 4}}{9} = \frac{{ - 4}}{9};\end{array}\)

 \(\begin{array}{l} + )\frac{{ - 3}}{7} + \frac{7}{{15}} + \frac{4}{{ - 7}} + \frac{8}{{15}} + \frac{2}{3}\\ = \frac{{ - 3}}{7} + \frac{7}{{15}} + \frac{{ - 4}}{7} + \frac{8}{{15}} + \frac{2}{3}\\ = \left( {\frac{{ - 3}}{7} + \frac{{ - 4}}{7}} \right) + \left( {\frac{7}{{15}} + \frac{8}{{15}}} \right) + \frac{2}{3}\\ = \frac{{ - 7}}{7} + \frac{{15}}{{15}} + \frac{2}{3}\\ = ( - 1) + 1 + \frac{2}{3} = 0 + \frac{2}{3} = \frac{2}{3}.\end{array}\)

 Do đó  \(\frac{-4}{9}<\frac{x}{9}\le \frac{2}{3}\) hay \(\frac{-4}{9}<\frac{x}{9}\le \frac{6}{9}\), từ đó suy ra \(-4<x\le 6\)

Vậy \(x\in \text{ }\!\!\{\!\!\text{ }-3;-2;-1;\,\,0;\,\,1;\,\,2;\,\,3;\,\,4;\,\,5;\,\,6\}\)