x + \(\frac{2}{3}\)= \(\frac{9}{11}\)
x - \(\frac{3}{10}\)=\(\frac{4}{15}\)
X nhân \(\frac{1}{7}\)=\(\frac{5}{6}\)
x : \(\frac{3}{5}\)= \(\frac{1}{6}\)
K
Khách
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15 tháng 3 2019
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
18 tháng 3 2020
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
18 tháng 3 2020
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)
22 tháng 3 2020
a, Ta có : \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}=\frac{x+7}{15}\)
=> \(3\left(2x-1\right)-5\left(x-2\right)=x+7\)
=> \(6x-3-5x+10-x-7=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
b, Ta có : \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
=> \(\frac{3\left(x+3\right)}{6}-\frac{2\left(x-1\right)}{6}=\frac{x+5}{6}+\frac{6}{6}\)
=> \(3\left(x+3\right)-2\left(x-1\right)=x+5+6\)
=> \(3x+9-2x+2-x-5-6=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
=> \(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)
=> \(4\left(x+5\right)+3\left(x+12\right)-5\left(x-2\right)=2x+66\)
=> \(4x+20+3x+36-5x+10-2x-66=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
HT
25 tháng 7 2017
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(x+\frac{2}{3}=\frac{9}{11}\)
\(x=\frac{9}{11}-\frac{2}{3}\)
\(x=\frac{5}{33}\)
\(x-\frac{3}{10}=\frac{4}{15}\)
\(x=\frac{3}{10}-\frac{4}{15}\)
\(x=\frac{1}{10}\)
\(x\times\frac{1}{7}=\frac{5}{6}\)
\(x=\frac{5}{6}:\frac{1}{7}\)
\(x=\frac{35}{6}\)
\(x:\frac{3}{5}=\frac{1}{6}\)
\(x=\frac{3}{5}.\frac{1}{6}\)
\(x=\frac{1}{10}\)
+, \(x+\frac{2}{3}=\frac{9}{11}\)
\(x=\frac{9}{11}-\frac{2}{3}=\frac{5}{33}\)
+, \(x-\frac{3}{10}=\frac{4}{15}\)
\(x=\frac{4}{15}+\frac{3}{10}=\frac{17}{30}\)
+, \(x\cdot\frac{1}{7}=\frac{5}{6}\)
\(x=\frac{5}{6}:\frac{1}{7}=\frac{5}{6}.\frac{7}{1}=\frac{35}{6}\)
+,\(x:\frac{3}{5}=\frac{1}{6}\)
\(x=\frac{1}{6}\cdot\frac{3}{5}=\frac{3}{30}=\frac{1}{10}\)