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\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)
\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)
=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
Do 7/15 < x/40 => 7.40 < 15.x
=> 280 < 15.x
=> 18 < x (1)
Do x/40 < 8/15 => x.15 < 8.40
=> x.15 < 320
=> x < 22 (2)
Từ (1) và (2), do x là số tự nhiên => x thuộc {19 ; 20 ; 21}
a) \(\frac{x}{7}+\frac{1}{14}=-\frac{1}{y}\)
\(\Rightarrow\frac{2x}{14}+\frac{1}{14}=\frac{-1}{y}\)
\(\Rightarrow\frac{2x+1}{14}=\frac{-1}{y}\)
\(\Rightarrow\left(2x+1\right).y=\left(-1\right).14=\left(-14\right)\)
Ta có bảng sau :
| 2x + 1 | 1 | -1 | 14 | -14 | 2 | -2 | 7 | -7 |
| 2x | 0 | -2 | 13 | -15 | 1 | -3 | 6 | -8 |
| x | 0 | -1 | \(\frac{13}{2}\) | \(\frac{-15}{2}\) | \(\frac{1}{2}\) | \(\frac{-3}{2}\) | 3 | -4 |
| y | -14 | 14 | -1 | 1 | -7 | 7 | -2 | 2 |
Vậy \(\left(x;y\right)\in\left\{\left(-1;14\right),\left(3;-2\right),\left(0;-14\right),\left(-4;2\right)\right\}\)
b) \(\frac{x}{9}+-\frac{1}{6}=-\frac{1}{y}\)
\(\Rightarrow\frac{2x}{18}+\frac{-3}{18}=\frac{-1}{y}\)
\(\Rightarrow\frac{2x-3}{18}=\frac{-1}{y}\)
\(\Rightarrow\left(2x-3\right).y=\left(-1\right).18=\left(-18\right)\)
Ta có bảng :
| 2x - 3 | 1 | -1 | 18 | -18 | 3 | -3 | 6 | -6 | 9 | -9 | -2 | 2 | ||||
| 2x | 4 | 2 | 21 | -15 | 6 | 0 | 9 | -3 | 12 | -6 | 1 | 5 | ||||
| x | 2 | 1 | \(\frac{21}{2}\) | \(\frac{-15}{2}\) | 3 | 0 | \(\frac{9}{2}\) | \(\frac{-3}{2}\) | 6 | -3 | \(\frac{1}{2}\) | \(\frac{5}{2}\) | ||||
| y | -18 | 18 | -1 | 1 | -6 | 6 | -3 | 3 | -2 | 2 | 9 | -9 |
Vậy \(\left(x;y\right)\in\left\{\left(2;-18\right),\left(1;18\right),\left(3;-6\right),\left(0;6\right),\left(6;-2\right),\left(-3,2\right)\right\}\)
a) \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)< \(x\) < \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
Ta có: \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)
=\(\frac{-40}{15}+\frac{21}{15}+\frac{-71}{15}\)
=\(\frac{-90}{15}\)
=\(-6\)
Ta có: \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
=\(\frac{-26}{14}+\frac{19}{14}+\frac{-49}{14}\)
=\(\frac{-56}{14}\)
=\(-4\)
=> \(-6\)< \(x\)<\(-4\)
=> \(x=-5\)
b)\(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)< \(\frac{x}{9}\)<\(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
Ta có: \(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)
=\(\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}+\frac{-11}{31}\right)+\frac{-4}{9}\)
=\(1+\left(-1\right)+\frac{-4}{9}\)
=\(0+\frac{-4}{9}\)
=\(\frac{-4}{9}\)
Ta có: \(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
=\(\frac{-3}{7}+\frac{7}{15}+\frac{-4}{7}+\frac{8}{15}+\frac{2}{3}\)
=\(\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{2}{3}\)
=\(\left(-1\right)+1+\frac{2}{3}\)
=\(0+\frac{2}{3}\)
=\(\frac{2}{3}\)
=> \(\frac{-4}{9}\)< \(\frac{x}{9}\)<\(\frac{2}{3}\)
=
=> \(\frac{-4}{9}\)<\(\frac{x}{9}\)<\(\frac{6}{9}\)
=> \(-4\)< \(x\)<\(6\)
=>\(x\in\left\{-3;-2;-1;0;1;2;3;4;5\right\}\)