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\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
\(\frac{24}{-12}\) = \(\frac{x}{5}\) = \(\frac{-y}{3}\)
- 2 = \(\frac{x}{5}\) = \(\frac{-y}{3}\)
\(x=5.\left(-2\right)\) = -10
y = -2.3:(-1) = -6:(-1) = 6
Vậy (\(x;y\)) = (-10; 6)
a/ \(-4\frac{3}{5}\cdot2\frac{4}{23}=\left(-\frac{23}{5}\right)\cdot\frac{50}{23}=-10\)
\(-2\frac{3}{5}\cdot1\frac{6}{15}=\left(-\frac{13}{5}\right)\cdot\frac{7}{5}=-\frac{91}{25}=-\frac{3}{64}\)
Mà \(-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}\cdot1\frac{6}{15}\)
=> \(-10\le x\le-3.64\)
=> \(x=\left\{-10;-9;-8;-7;-6;-5;-4;-3\right\}\)
<_ là dấu \(\ge\)??