Tìm x \(\in Z\)
\(\frac{x}{3}=\frac{24}{x+1}\), \(\frac{-10}{3}+\frac{-2}{3}\le x< \frac{-2}{7}+\frac{-5}{7}\)
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Có: \(4.\frac{-3}{10}\le x\le\frac{3}{11}.\frac{11}{30}\Rightarrow\frac{-6}{5}\le x\le\frac{1}{10}\)
\(\Rightarrow-\frac{12}{10}\le x\le\frac{1}{10}\) mà x là số nguyên \(\Rightarrow x=-1\)
\(\frac{3}{x-5}=-\frac{4}{x+2}\)
\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
\(\frac{x}{-2}=-\frac{8}{x}\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
\(-\frac{2}{x}=\frac{y}{3}\)
\(\Leftrightarrow xy=-6\)
\(\Leftrightarrow x;y\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Xét bảng
x | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -1 |
y | -6 | 6 | -3 | 3 | -2 | 2 | -1 | 6 |
Vậy.................
\(\frac{2x-9}{240}=\frac{39}{80}\)
\(\Leftrightarrow2x-9=\frac{240.39}{80}\)
\(\Leftrightarrow2x-9=117\)
\(\Leftrightarrow2x=126\)
\(\Leftrightarrow x=63\)
\(\frac{2}{3}\) .\(\frac{3}{4}\)\(\le\)\(\frac{x}{18}\) \(\le\)\(\frac{7}{3}\).\(\frac{1}{3}\)
\(\frac{1}{2}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{9}{18}\le\frac{x}{18}\le\frac{14}{18}\)
\(\Rightarrow x\in\){9:10;11;12;13;14}
\(\frac{2}{3}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(\frac{2}{3}.\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\frac{1}{3}\)
\(\frac{2}{3}.\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)
Vậy \(x\in\left\{11;12;13\right\}\)
\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)
\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)
2 câu sau tự làm nha
\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)
\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)
=> \(x\in\left\{-2;-1;0;1;2\right\}\)
\(\frac{3}{7}\cdot15\cdot\frac{1}{3}+\frac{3}{7}\cdot5\cdot\frac{2}{5}\le x\le\left(3\frac{1}{2}:7-6\frac{1}{2}\right)\cdot\left(-2\frac{1}{3}\right)\)
\(\Leftrightarrow\frac{15}{7}+\frac{6}{7}\le x\le-6\cdot\frac{-5}{3}\)
\(\Leftrightarrow3\le x\le10\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
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